atcoder#AGC019E. [AGC019E] Shuffle and Swap

[AGC019E] Shuffle and Swap

Score : 17001700 points

Problem Statement

You have two strings A=A1A2...AnA = A_1 A_2 ... A_n and B=B1B2...BnB = B_1 B_2 ... B_n of the same length consisting of 00 and 11. The number of 11's in AA and BB is equal.

You've decided to transform AA using the following algorithm:

  • Let a1a_1, a2a_2, ..., aka_k be the indices of 11's in AA.
  • Let b1b_1, b2b_2, ..., bkb_k be the indices of 11's in BB.
  • Replace aa and bb with their random permutations, chosen independently and uniformly.
  • For each ii from 11 to kk, in order, swap AaiA_{a_i} and AbiA_{b_i}.

Let PP be the probability that strings AA and BB become equal after the procedure above.

Let Z=P×(k!)2Z = P \times (k!)^2. Clearly, ZZ is an integer.

Find ZZ modulo 998244353998244353.

Constraints

  • 1A=B10,0001 \leq |A| = |B| \leq 10,000
  • AA and BB consist of 00 and 11.
  • AA and BB contain the same number of 11's.
  • AA and BB contain at least one 11.

Partial Score

  • 12001200 points will be awarded for passing the testset satisfying 1A=B5001 \leq |A| = |B| \leq 500.

Input

Input is given from Standard Input in the following format:

AA

BB

Output

Print the value of ZZ modulo 998244353998244353.

1010
1100
3

After the first two steps, a=[1,3]a = [1, 3] and b=[1,2]b = [1, 2]. There are 44 possible scenarios after shuffling aa and bb:

  • a=[1,3]a = [1, 3], b=[1,2]b = [1, 2]. Initially, A=1010A = 1010. After swap(A1A_1, A1A_1), A=1010A = 1010. After swap(A3A_3, A2A_2), A=1100A = 1100.
  • a=[1,3]a = [1, 3], b=[2,1]b = [2, 1]. Initially, A=1010A = 1010. After swap(A1A_1, A2A_2), A=0110A = 0110. After swap(A3A_3, A1A_1), A=1100A = 1100.
  • a=[3,1]a = [3, 1], b=[1,2]b = [1, 2]. Initially, A=1010A = 1010. After swap(A3A_3, A1A_1), A=1010A = 1010. After swap(A1A_1, A2A_2), A=0110A = 0110.
  • a=[3,1]a = [3, 1], b=[2,1]b = [2, 1]. Initially, A=1010A = 1010. After swap(A3A_3, A2A_2), A=1100A = 1100. After swap(A1A_1, A1A_1), A=1100A = 1100.

Out of 44 scenarios, 33 of them result in A=BA = B. Therefore, P=3P = 3 / 44, and Z=3Z = 3.

01001
01001
4

No swap ever changes AA, so we'll always have A=BA = B.

101010
010101
36

Every possible sequence of three swaps puts the 11's in AA into the right places.

1101011011110
0111101011101
932171449