Problem:
Score : 1700 points

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.

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.

Partial Score

1200 points will be awarded for passing the testset satisfying 1 \leq |A| = |B| \leq 500.

Input

Input is given from Standard Input in the following format:

A
B

Output

Print the value of Z modulo 998244353.

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

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
#define syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<string> vs;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<pll> vpll;
typedef pair<int,P> pip;
typedef vector<pip> vip;
const int inf=1<<30;
const ll INF=1ll<<60;
const double pi=acos(-1);
const double eps=1e-8;
const ll mod=998244353;
const int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};

ll Pow(ll n,ll p){
	ll r=1;
	for(;p>0;p>>=1){
		if(p&1) r=(r*n)%mod;
		n=(n*n)%mod;
	}
	return r;
}

const int M=100005;
ll F[M];

void Init(){
	F[0]=1;
	for(int i=1;i<M;i++) F[i]=F[i-1]*i%mod;
}

ll Div(ll n,ll m){
	return n*Pow(m,mod-2)%mod;
}

ll nCk(ll n,ll k){
	return Div(F[n],F[n-k]*F[k]%mod);
}

const int N=600;
int n;
string s,t;
vvl dp(N,vl(N,-1));

ll f(ll i,ll j){
	if(dp[i][j]!=-1) return dp[i][j];
	if(!i&&!j) return dp[i][j]=1;
	ll res=0;
	if(i>1) res=i*(i-1)*f(i-2,j+2)%mod;
	if(i) (res+=i*(j+1)*f(i-1,j))%=mod;
	if(j>1) (res+=j*(j-1)/2*f(i,j-2))%=mod;
	return dp[i][j]=res;
}

int main(){
	Init();
	cin>>s>>t;
	n=s.size();
	int A=0,B=0,C=0;
	for(int i=0;i<n;i++){
		if(s[i]=='1'&&t[i]=='1') A++;
		if(s[i]=='1'&&t[i]=='0') B++;
		if(s[i]=='0'&&t[i]=='1') C++;
	}
	if(B!=C){
		cout<<0<<endl;
		return 0;
	}
	assert(0);
	if(!B){
		cout<<f(A,0)<<endl;
		return 0;
	}
	ll res=0;
	for(int i=0;i<=A;i++){
		(res+=nCk(A+B,i)*F[B]%mod*F[B]%mod*F[A-i]%mod*nCk(A+B-i-1,B-1)%mod*f(i,0))%=mod;
	}
	cout<<res<<endl;
}