Two positive integers *a* and *b* have a sum of *s* and a bitwise XOR of *x*. How many possible values are there for the ordered pair (*a*, *b*)?
The first line of the input contains two integers *s* and *x* (2 ≤ *s* ≤ 10<sup class="upper-index">12</sup>, 0 ≤ *x* ≤ 10<sup class="upper-index">12</sup>), the sum and bitwise xor of the pair of positive integers, respectively.
In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).

In the second sample, the only solutions are (1, 2) and (2, 1).
9 5
4
3 3
2
5 2
0
6 0
1
549755813887 549755813887
549755813886
2 0
1
2 2
0
433864631347 597596794426
0
80 12
4
549755813888 549755813886
274877906944
643057379466 24429729346
2048
735465350041 356516240229
32768
608032203317 318063018433
4096
185407964720 148793115916
16384
322414792152 285840263184
4096
547616456703 547599679487
68719476736
274861129991 274861129463
34359738368
549688705887 549688703839
34359738368
412182675455 412182609919
68719476736
552972910589 546530328573
17179869184
274869346299 274869346299
8589934590
341374319077 341374319077
134217726
232040172650 232040172650
65534
322373798090 322373798090
1048574
18436 18436
6
137707749376 137707749376
30
9126813696 9126813696
6
419432708 419432708
62
1839714 248080
128
497110 38
8
1420572 139928
64
583545 583545
4094
33411 33411
30
66068 66068
14
320 320
2
1530587 566563
256
1988518 108632
128
915425594051 155160267299
0
176901202458 21535662096
0
865893190664 224852444148
32768
297044970199 121204864
0
241173201018 236676464482
0
1582116 139808
0
1707011 656387
0
169616 132704
32
2160101 553812
0
1322568 271816
0
228503520839 471917524248
0
32576550340 504864993495
0
910648542843 537125462055
0
751720572344 569387893618
0
629791564846 602334362179
0
1000000000000 1000000000000
8190
1000000000000 999999999999
0
1000000000000 4
0
1000000000000 4096
2
3 1
0
2097152 0
1
40 390
0
22212 39957
0
128 36
0
14 4
0
6 2
0
43 18467
0
7 1
0
7 5
0
251059 79687
0
17 7
0
4 6
0
2 4
0
3 7
0
from sys import stdin,stdout
r,w = stdin.readline,stdout.write
s,x = map(int,r().split());andShft = s-x
w('%i\n' %(0 if andShft&1 or s<x or x&(andShft>>1) else 2**(f'{x:b}'.count('1')) - (2 if s is x else 0)))