IEEE C80216m-09_1814r1
Project
IEEE 802.16 Broadband Wireless Access Working Group <http://ieee802.org/16>
Title
Proposed “Resource Allocation” field in Assignment A-MAP IEs
Date
Submitted
2009-08-24
Source(s)
Yanfeng Guan Xiangyu Liu, Ying Liu,
Huiying Fang, Bo Sun
[email protected]
ZTE Corporation
Comment on “DRAFT Amendment to IEEE Standard for Local and metropolitan area networks,
P802.16m/D1”
Re:
15.3.6 Downlink Control Structure & 15.3.8 Uplink Control Structure
Abstract
This contribution proposes text specifying the structure and interpretation of the “Resource
Allocation” filed in the A-AMAP IEs for the 5, 10 & 20 MHz in some special cases.
Purpose
To be discussed and adopted by TGm for the 802.16m amendment.
Notice
Release
Patent
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Proposed “Resource Allocation” field in Assignment A-MAP IEs
Yanfeng Guan, Xiangyu Liu, Ying Liu, Huiying Fang, Bo Sun
ZTE Corporation
1) Introduction
In current 802.16m draft amendment [1], four types of Assignment A-MAP IEs are defined and used for
dynamic scheduling. They are DL/UL Basic Assignment A-MAP IEs for an allocation of contiguous LRUs and
DL/UL Sub-band Assignment A-MAP IEs for an allocation of subband-based LRUs in DL/UL subframes. For
these four IEs, the main difference is the interpretation for the “Resource Index” field.
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IEEE C80216m-09_1814r1
2) Problem
2.1) Resource Indexing in Basic Assignment A-MAP IE
RI (Resource Index) field indicates the location and size of one instance of a resource allocation for an AMS.
Although some good design considerations in [2] were taken into account, but there are still some other issues
needed to be considered.
For RA field in DL/UL Basic Assignment A-MAP IEs, the pre-processing is derivation of the mapping between
LRU index and physical PRU index. Mapping CRU/DRU to LRU is defined as:
CRU FP [k  Si 1 ],
LRU [k ]  
for Si 1  k  (LCRU , FPi  Si 1 ) 
i
DRU FP [k  ( LCRU , FPi  Si 1 )],

i

where 0  k 
m 3

m0
FPSm . and Si 
mi

m0
for (LCRU , FPi

 , for 0  i  3,
 Si 1 )  k  Si 

FPSm , 0  i  3
Obviously, the derivate LRU index is according Frequency Partition Number, not the DRUs and CRUs, as a
result of that, all the CRUs or DRUs are non-contiguous in logical index, and then many allocation granularities
can’t be implemented when FPCT is more than 1.
We can see an example regarding the resource mapping in D1. The following figure is a general instance for the
downlink resource mapping. Assuming that FP0 is Reuse 1 Region and FP1 is high boosted Reuse 3 Region, the
power used in FP2 is same or similar to FP3 because both they are low boosted Reuse 3 Regions. However,
ABS can’t assign the resource in FP2 and FP3 simultaneously in one Basic Assignment A-MAP IE because the
resource indexing method only supports the contiguous allocation by Basic IE.
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IEEE C80216m-09_1814r1
PRUFP0
PRUSB
PRU
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
0
1
2
3
8
9
10
11
16
17
18
19
24
25
26
27
32
33
34
35
40
41
42
43
4
5
6
7
PRUMB
12
13
14
15
20
21
22
23
28
29
30
31
36
37
38
39
44
45
46
47
PRUSB
0
1
2
3
8
9
10
11
16
17
18
19
24
25
26
27
32
33
34
35
40
41
42
43
4
5
6
7
PPRUMB
12
20
28
36
44
13
21
29
37
45
14
22
30
38
46
15
23
31
39
47
0
1
2
3
12
20
28
36
44
13
21
29
PRUFP1
8
9
10
11
16
17
18
19
37
45
14
22
PRUFP2
24
25
26
27
32
33
34
35
30
38
46
15
PRUFP3
40
41
42
43
4
5
6
7
23
31
39
47
CRU(FP0)
0
1
2
3
20
21
DRU(FP0)
12
28
36
44
13
29
CRU(FP1)
8
9
10
11
16
17
18
19
DRU(FP1)
37
45
14
22
CRU(FP2)
24
25
26
27
32
33
34
35
DRU(FP2)
30
38
46
15
CRU(FP3)
40
41
42
43
4
5
6
7
DRU(FP3)
23
31
39
47
Figure - CRU/DRU allocation for BW=10 MHz, KSB = DSAC = 7, FPCT = 4, FPSi = 12 (for i≥0), DFPSC = 2,
DCASSB,0 = 1, DCASMB,0 = 2, and DCASi=2
This problem is not only a special case, actually, it will be present for many resource mapping configurations in
DL/UL Resource Mapping Configurations.
2.2) Resource Indexing in Subband Assignment A-MAP IE
For RA field in DL/UL Subband Assignment A-MAP IEs, the indexing method means that the ABS can use two
different resource allocation granularities (a subband and a half subband) for the different bandwidth or different
number of subband. But there are some problems:
1) There is a mistake in pre-processing part, the real number of subband for frequency partition i (i>0) is
indicated by LSB,FPi, but CASSB,FPi.
1) When Y is equal to 6 and ITF is present, only 6 bits are used to indicate the resource allocation. So this will
lead that many cases can’t be indicated because we can use 9 bits available.
2) For 20MHz, the method is too complicated. In addition, Table for triplets of non-contiguous sub-band indices
is TBD.
In this contribution, we propose some modifications on the “Resource Allocation” field in DL/UL
3
IEEE C80216m-09_1814r1
Basic/Subband-Assignment A-MAP IE.
3) References
[1] "DRAFT Amendment to IEEE Standard for Local and metropolitan area networks Part 16: Air Interface for Broadband
Wireless Access Systems Advanced Air Interface", P802.16m/D1 July 2009.
[2] IEEE 802.16m-07/002r8, “IEEE 802.16m System Requirements Document”
[3] IEEE 802.16m-08/003r7, “IEEE 802.16m System Description Document”
[4] IEEE 802.16m-08/043, “Style guide for writing the IEEE 802.16m amendment”
[5] IEEE 802.16m-09/1327r1, “Proposed Changes to the DL and UL A-MAP Information Element Types in AWD ”
[6] IEEE 802.16m-09/1331r2, “Proposed AWD text specifying the “Resource Allocation” field in the A-MAP IEs AWD ”
[7] IEEE 802.16m-09/1334r1, “AWD Proposal for DL/UL Resource Indexing in IEEE 802.16m ”
[8] IEEE 802.16m-09/1516r “Proposed “Resource Allocation” Field in Basic Assignment A-MAP IE in IEEE802.16m ”
4) Proposed Text
Proposed Text for Remedy 1:
Modify “Derivation of the mapping between LRU index and physical PRU index” part on page 326 in the
15.3.6.5.4.2 DL basic assignment A-MAP IE as followed:
----------------------------------------------------- Proposed Text Starts -------------------------------------------------Derivation of the mapping between LRU index and physical PRU index:
For each frequency partition i, tThe total number of CLRUs &and DLRUs up-to and including that partition i is
calculated as
Si 
mi

m0
SCRU ,i 
SDRU ,i 
FPSm , 0  i  3
m i
L

m0
CRU，FPm
, 0  i  FPCT 1
(200)
, 0  i  FPCT 1
(200a)
m i
L
DRU，FPm
m0
(200)
CLRU [ j ]  CRU FPi [ j  SCRU ,i 1 ], SCRU ,i 1  j  SCRU ,i ), 0  i  FPCT 1
(200b)
DLRU [ j ]  DRU FPi [ j  S DRU ,i 1 ], SDRU ,i 1  j  SDRU ,i ), 0  i  FPCT 1
(200c)
CRU FP [k  Si 1 ],
LRU [k ]  
i

i
 DRU FP [k  ( LCRU , FPi  Si 1 )],
for Si 1  k  (LCRU , FPi  Si 1 ) 
for (LCRU , FPi
4

 , for 0  i  3,
 Si 1 )  k  Si 

(201)
IEEE C80216m-09_1814r1
0  k  SCRU , FPCT 1
CLRU [k ],


(201)
LRU [k ]  

DLRU
[
k

S
],
S

k

S

S
CRU
,
FPCT

1
CRU
,
FPCT

1
CRU
,
FPCT

1
DRU
,
FPCT

1




with 0  k 
m 3

m0
with 0  k 
FPSm .
m  FPCT-1

m0
FPSm .
The mapping from CRUFPi and DRUFPi to the physical PRU indices (and vice-versa) is as specified in Section
15.3.5 and Section 15.3.8.
The LRU index, L obtained from the RI field refers to the LRU[ ] defined above.
------------------------------------------------------ Proposed Text Ends --------------------------------------------------
Proposed Text for Remedy 2:
Modify indexing method part on page 311 in the 15.3.6.5.1 DL Sub-band Assignment A-MAP IE as followed:
------------------------------------------------------ Proposed Text Starts -------------------------------------------------15.3.6.5.1. DL Sub-band Assignment A-MAP IE
……
15.3.6.5.1.1 Pre-Processing
The mapping between the LRU indices and the physical PRU indices (and vice-versa) may be derived as
described in Section 15.3.6.5.2.2 (“DL Basic Assignment A-MAP IE”).
Derivation of the mapping between the sub-band-CRU index and the physical PRU index: shall be done as
follows.
The total number of sub-bands over all partitions may be is calculated as
m 3

Y
m0
Y
DCASSB,m
m  FPCT 1

m0
(Eq. 001)
LSB，FP m / N1
For each frequency partition i, tThe total number of sub-band-CRUs up-to and including that partition may be i
is calculated as
Xi 
mi

m0
N1.DCASSB,m , 0  i  3.
5
Xi 
IEEE C80216m-09_1814r1
m i
L

m0
SB，FPm
/ N1 , 0  i  FPCT 1
(Eq. 002)
Then, the logical sub-band CRUs are indexed as
SBCRU [k ]  CRU FP [k  X i 1],
i
for Xi 1  k  ( N1.DCAS SB,i  X i 1), with 0  i  3,
SBCRU [k ]  CRU FP [k  X i 1 ], X i 1  j  X i ), 0  i  FPCT 1
(Eq. 003)
i
with 0  k  N1.Y .
The mapping from the CRUFPi to the physical PRU indices (and vice-versa) is as specified in Section 15.3.5 and
Section 15.3.8.
The LRU & SBCRU indices indicated by the RA field refer to the LRU[] and the SBCRU[] defined above
15.3.6.5.1.2 Nominal Channel Bandwidth = 5 MHz
For a 5 MHz system:
1) A single instance of a resource allocation shall be made using a single IE.
2) The AMS shall interpret the RA field as defined by Table 668.
Table 668: Interpretation of the RA Field in a 5 MHz system
Total # of
sub-bands
over all
frequency
partitions,
Y
Y=6
RA Field Interpretation
The 2 MSB bits of the RA field are denoted as the “Indication Type field” (ITF).
The 6 9 LSB bits of the RA field are denoted as the “Resource Indexing Field” (RIF).
Denote RIF[j] as the jth bit position in the RIF, 0 <= j < 6 9, with j = 0 being the LSB.
ITF == 00
Each of the RIF[j], 0 <= j < 6, indicates the allocation or non-allocation of all N1 (= 4) CRUs within a
particular sub-band.
RIF[ j ]  1  The 4 CRUs with indices k such that  SBCRU [k ]  = j have been allocated.


N1


RIF[ j ]  0  The 4 CRUs with indices k such that  SBCRU [k ]  = j have been not allocated.


N1


ITF == 01
Each of the RIF[j], 0 <= j < 69, indicates the allocation or non-allocation of the 1st or last 2 CRUs
within a particular sub-band.
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IEEE C80216m-09_1814r1
 SBCRU [k ] 
 =2j have been allocated.
 ( N1 / 2) 
RIF[j] = 1  The 2 CRUs with indices k such that 
 SBCRU [k ] 
 =2j have been not allocated.
 ( N1 / 2) 
RIF[j] = 0  The 2 CRUs with indices k such that 
ITF == 10
Each of the RIF[j], 0 <= j < 69, indicates the allocation or non-allocation of the last 2 CRUs within a
particular sub-band.
 SBCRU [k ]   SBCRU [23  k ] 
 
 =2j+1
( N1 / 2)
 ( N1 / 2)  

RIF[j] = 1  The 2 CRUs with indices k such that 
have been allocated.
 SBCRU [k ]   SBCRU [23  k ] 
 
 =2j+1
( N1 / 2)
 ( N1 / 2)  

RIF[j] = 0  The 2 CRUs with indices k such that 
Y <= 5
have been not allocated.
ITF == 11 Reserved
Each of the j bit-positions in the RA field, 0 <= j < 11, indicates the allocation or non-allocation of 2 CRUs within a
particular sub-band.
Position j in RA corresponds to the 2 CRUs with indices k such that
 SBCRU [k ] 
 N / 2  =j.


1
Depending on the value of Y, not all j may be relevant.
 SBCRU [k ] 
 = j have been allocated.
 N1 / 2 
RA[j] = 1  The 2 CRUs with indices k such that 
 SBCRU [k ] 
 = j have been not allocated.
 N1 / 2 
RA[j] = 0  The 2 CRUs with indices k such that 
15.3.6.5.1.3 Nominal Channel Bandwidth = 10 MHz
For a 10 MHz system a single instance of a resource allocation shall be made using a single IE. Furthermore when Y (the total
number of sub-bands over all frequency partitions) <= 11, the AMS shall interpret the bits in the Resource Allocation field as
indicated by Table 669 (the row corresponding to the particular value of Y in that table). In the case that Y = 12, the RA field
shall be interpreted as in the first row of Table 669.
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IEEE C80216m-09_1814r1
Table 669: Interpretation of the RA Field in a 10 MHz or a 20 MHz system, when Y <= 11 (Y is
the total number of sub-bands over all frequency partition)
Total # of
RA Field Interpretation
sub-bands
over all
frequency
partitions, Y
Y = 10 or 11
Each of the j bit-positions in the RA field, 0 <= j < 11, indicates the allocation or non-allocation of all N1 (= 4)
CRUs within a particular sub-band.
Denote RA[j] as the jth bit position in the RA Field, 0 <= j < 11, with j = 0 being the LSB.
 SBCRU [k ] 

 = j.
N


1
RA[j] corresponds to the subband, i.e., the 4 CRUs with indices k such that
Depending on the value of Y , not all j may be relevant.
RIF[j] = 1  The 4 CRUs with indices k such that
RIF[j] = 0  The 4 CRUs with indices k such that
Y = 6,7,8 or
9
 SBCRU [k ] 

 =j have been allocated.
N1


 SBCRU [k ] 

 =j have been not allocated.
N1


The 2 MSB bits of the RA field are denoted as the “Indication Type field” (ITF).
The Y LSB bits of the RA field are denoted as the “Resource Indexing Field” (RIF).
Denote RIF[j] as the jth bit position in the RIF, 0 <= j < Y, with j = 0 being the LSB.
ITF == 00 Each of the RIF[j], 0 <= j < Y, indicates the allocation or non-allocation of all N1 (= 4) CRUs within a
particular sub-band (out of the Y sub-bands).
 SBCRU [k ] 

 =j have been allocated.
N


1
 SBCRU [k ] 

 =j have been not allocated
N1


RIF[j] = 1  The 4 CRUs with indices k such that
RIF[j] = 0  The 4 CRUs with indices k such that
ITF == 01
Each of the RIF[j], 0 <= j < Y, indicates the allocation or non-allocation of the 1st 2 CRUs within a
particular sub-band.
RIF[j] = 1  The 2 CRUs with indices k such that
RIF[j] = 0  The 2 CRUs with indices k such that
ITF == 10
 SBCRU [k ] 
 N / 2  = 2j have been allocated.


1
 SBCRU [k ] 
 N / 2  = 2j have been not allocated.


1
Each of the RIF[j], 0 <= j < Y, indicates the allocation or non-allocation of the last 2 CRUs within a
particular sub-band.
 SBCRU [k ]   SBCRU [ N1 Y  k  1] 
 = 2j
 
( N1 / 2)

 ( N1 / 2)  
RIF[j] = 1  The 2 CRUs with indices k such that 
+1 have been allocated.
 SBCRU [k ]   SBCRU [ N1 Y  k  1] 
 = 2j
 
( N1 / 2)

 ( N1 / 2)  
RIF[j] = 0  The 2 CRUs with indices k such that 
+1 have been not allocated.
8
IEEE C80216m-09_1814r1
Y=6
The 2 MSB bits of the RA field are denoted as the “Indication Type field” (ITF).
The 9 LSB bits of the RA field are denoted as the “Resource Indexing Field” (RIF).
Denote RIF[j] as the jth bit position in the RIF, 0 <= j < 9, with j = 0 being the LSB.
ITF == 00
ITF == 01
Each of the RIF[j], 0 <= j < 6, indicates the allocation or non-allocation of all N1 (= 4) CRUs within a
particular sub-band.
RIF[j] = 1  The 4 CRUs with indices k such that
 SBCRU [k ] 

 =j have been allocated.
N1


RIF[j] = 0  The 4 CRUs with indices k such that
 SBCRU [k ] 

 =j have been not allocated
N1


Each of the RIF[j], 0 <= j < 9, indicates the allocation or non-allocation of the 1st or last 2 CRUs
within a particular sub-band.
 SBCRU [k ] 
 = j have been allocated.
 N 1/2 
RIF[j] = 1  The 2 CRUs with indices k such that 
 SBCRU [k ] 
 N / 2  = j have been not allocated.


1
RIF[j] = 0  The 2 CRUs with indices k such that
ITF == 10
Each of the RIF[j], 0 <= j < 9, indicates the allocation or non-allocation of the 1st or last 2 CRUs
within a particular sub-band.
 SBCRU [23  k ] 
 = j have been allocated.
N 1/2


RIF[j] = 1  The 2 CRUs with indices k such that 
 SBCRU [23  k ] 
 = j have been not allocated.
N 1/2


RIF[j] = 0  The 2 CRUs with indices k such that 
ITF == 11
Y <= 5
Reserved
Each of the j bit-positions in the RA field, 0 <= j < 11, indicates the allocation or non-allocation of 2 CRUs within a
particular sub-band.
Position j in RA corresponds to the 2 CRUs with indices k such that
 SBCRU [k ] 
 N / 2  =j.


1
Depending on the value of Y, not all j may be relevant.
 SBCRU [k ] 
 = j have been allocated.
 N1 / 2 
RA[j] = 1  The 2 CRUs with indices k such that 
 SBCRU [k ] 
 = j have been not allocated.
 N1 / 2 
RA[j] = 0  The 2 CRUs with indices k such that 
15.3.6.5.1.3 Nominal Channel Bandwidth = 20 MHz
For a 20 MHz system the following procedure shall be followed:
1) When Y SSB (total number of sub-bands over all frequency partitions) <= 11, a single instance of a resource allocation shall
be made using a single IE. In this case, the AMS shall interpret the bits in the Resource Allocation field as indicated by
Table 669 (the row corresponding to the particular value of Y in that table).
2) When Y (total number of sub-bands over all frequency partitions) > 11, a single instance of a resource allocation may be
made using a single or two IEs.
9
IEEE C80216m-09_1814r1
a) When an allocation is made using 2 IEs, the RA fields of the two IEs shall be concatenated to form a 22-bit field,
referred to as the Concatenated-RA field (C-RA field). The LSB of the RA field of the IE occurring last in the A-AMAP
region shall be interpreted as the LSB of the C-RA field. The AMS shall interpret the bits in the C-RA field as indicated by
Table 670 (the row corresponding to the particular value of Y in that table). The AMS may infer that two IEs refer to the
same instance of a resource allocation from the values of the ACID and SPID fields.
b) When an allocation is made using a single IE, the RA field shall be interpreted as in Table 671 & Table_TBD. These
tables map the value of the RA field to the indices of 2 non-contiguous sub-bands, and 3 non-contiguous sub-bands,
respectively.
Table 670— Interpretation of the C-RA Field in a 20 MHz system
10
IEEE C80216m-09_1814r1
Total # of
C-RA Field Interpretation
sub-bands
over all
frequency
partitions, Y
Y = 21,22,23
Each of the j bit-positions in the C-RA field, 0 <= j < 22, indicates the allocation or non-allocation of all N1 (= 4)
or 24
CRUs within a particular sub-band.
Denote RA[j] as the jth bit position in the C-RA Field, 0 <= j < 22, with j = 0 being the LSB.
 SBCRU [k ] 

 = j.
N1

RA[j] corresponds to the subband, i.e., the 4 CRUs with indices k such that 
Depending on the value of Y , not all j may be relevant.
RA[j] = 1  The 4 CRUs with indices k such that
RA[j] = 0  The 4 CRUs with indices k such that
1220 <= Y
<= 1220
 SBCRU [k ] 

 =j have been allocated.
N1


 SBCRU [k ] 

 =j have been not allocated.
N1


The 2 MSB bits of the C-RA field are denoted as the “Indication Type field” (ITF).
The Y LSB bits of the C-RA field are denoted as the “Resource Indexing Field” (RIF).
Denote RIF[j] as the jth bit position in the RIF, 0 <= j < Y, with j = 0 being the LSB.
ITF == 00 Each of the RIF[j], 0 <= j < Y, indicates the allocation or non-allocation of all N1 (= 4) CRUs within a
particular sub-band (out of the Y sub-bands).
 SBCRU [k ] 
 = j have been allocated.
N


1
RIF[j] = 1  The 4 CRUs with indices k such that 
 SBCRU [k ] 
 = j have been not allocated
N1


RIF[j] = 0  The 4 CRUs with indices k such that 
ITF == 01
Each of the RIF[j], 0 <= j < Y, indicates the allocation or non-allocation of the 1st 2 CRUs within a
particular sub-band.
 SBCRU [k ] 
 = 2j have been allocated.
 N1 / 2 
RIF[j] = 1  The 2 CRUs with indices k such that 
 SBCRU [k ] 
 = 2j have been not allocated.
 N1 / 2 
RIF[j] = 0  The 2 CRUs with indices k such that 
ITF == 10
Each of the RIF[j], 0 <= j < Y, indicates the allocation or non-allocation of the last 2 CRUs within a
particular sub-band.
 SBCRU [k ]   SBCRU [ N1 Y  k  1] 
 =
 
(
N
/
2)
(
N
/
2)




1
1
RIF[j] = 1  The 2 CRUs with indices k such that 
2j+1 have been allocated.
 SBCRU [k ]   SBCRU [ N1 Y  k  1] 
 = 2j+1
 
( N1 / 2)

 ( N1 / 2)  
RIF[j] = 0  The 2 CRUs with indices k such that 
ITF == 11
have been
Reserved
Table 671 maps the value of the 11-bit RA to a pair of non-contiguous sub-band indices.
11
IEEE C80216m-09_1814r1
Table 671—Mapping of decimal values of the 11-bit RA (for 0 <= RA value < 254) to a pair of noncontiguous sub-band indices, for a 10 MHz system
1st SBI
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
2nd SBI
0
1
2
3
4
5
6
7
8
9
10
0
9
21
22
42
43
62
63
81
82
99
100
116
117
132
133
147
148
161
162
174
175
11
186
187
12
197
198
13
207
208
14
216
217
15
224
225
16
231
232
17
237
238
18
242
243
19
246
247
20
249
250
21
251
252
In Table 671, the decimal value of the RA field increase from left to right, and then from top to bottom; values at the beginning
and end of each row are shown in the table. The shaded cells are not used to make interpretations. For a given value of the RA,
the mapping using that value is given by the indices of the two sub-bands as indicated in the “1st sub-band index” row and the
“2nd sub-band index” column. As an example, RA == 00001001 (Decimal value 9) maps to the pair of sub-bands indices {0,
 LRU [k ] 
 = 0 or 11.
 N1 
11}. Then, the allocation is the set of LRUs with index k such that 
Table Table_RA_5 translates values of the 11-bit RA from 253 to 2047 to triplets of non-contiguous sub-band indices. This
table is TBD.
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12