Problem Statement

You have two strings A = A_1 A_2 ... A_n and B = B_1 B_2 ... B_n of the same length consisting of 0 and 1.
The number of 1's in A and B is equal.
You've decided to transform A using the following algorithm:

• Let a_1, a_2, ..., a_k be the indices of 1's in A.
• Let b_1, b_2, ..., b_k be the indices of 1's in B.
• Replace a and b with their random permutations, chosen independently and uniformly.
• For each i from 1 to k, in order, swap A_{a_i} and A_{b_i}.

Let P be the probability that strings A and B become equal after the procedure above.
Let $Z = P \times (k$!)^2. Clearly, Z is an integer.
Find Z modulo 998244353.

Input

Input is given from Standard Input in the following format:
A
B

Output

Print the value of Z modulo 998244353.

Constraints

• $1 \leq |A| = |B| \leq 10,000$
• A and B consist of 0 and 1.
• A and B contain the same number of 1's.
• A and B contain at least one 1.

Sample Input 1

1010
1100

Sample Output 1

3

After the first two steps, a = [1, 3] and b = [1, 2]. There are 4 possible scenarios after shuffling a and b:

• a = [1, 3], b = [1, 2]. Initially, A = 1010. After swap(A_1, A_1), A = 1010. After swap(A_3, A_2), A = 1100.
• a = [1, 3], b = [2, 1]. Initially, A = 1010. After swap(A_1, A_2), A = 0110. After swap(A_3, A_1), A = 1100.
• a = [3, 1], b = [1, 2]. Initially, A = 1010. After swap(A_3, A_1), A = 1010. After swap(A_1, A_2), A = 0110.
• a = [3, 1], b = [2, 1]. Initially, A = 1010. After swap(A_3, A_2), A = 1100. After swap(A_1, A_1), A = 1100.

Out of 4 scenarios, 3 of them result in A = B. Therefore, P = 3 / 4, and Z = 3.

Sample Input 2

01001
01001

Sample Output 2

4

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

Sample Input 3

101010
010101

Sample Output 3

36

Every possible sequence of three swaps puts the 1's in A into the right places.

Sample Input 4

1101011011110
0111101011101

Sample Output 4

932171449

见题面 Input 部分

见题面 Output 部分