#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;
const int MAXN = 1e3+7;
const long long MOD = 998244353;
int n,m;
long long dp[MAXN][MAXN][2],f[MAXN][MAXN][2];
long long siz[MAXN][MAXN][2];
int sza[MAXN],szb[MAXN];

int main()
{
	string a,b;
	cin >> a >> b;
	n = a.length();m = b.length();
	a = " " + a;
	b = " " + b;
	int u,v;
	long long ans = 0;
	for(int i = 0;i<=n;i++) 
		for(int j = 0;j<=m;j++) 
			siz[i][j][0] = siz[i][j][1] = 1;
	for(int i = n;i>=0 ;i--){
		for(int j = m;j>=0 ;j--){
			if(i && a[i] != a[i-1]) {
				//dp[i][j][0] += dp[i-1][j][0];
				//dp[i][j][0]表式a串取1到i,b串取1到j,要使merge串的最后一位是a串的最后一位的构造方案数 
				siz[i-1][j][0] += siz[i][j][0];
				siz[i-1][j][0] %= MOD;
			} 
			if(i && a[i] != b[j]){
				//dp[i][j][0] += dp[i-1][j][1];
				siz[i-1][j][1] += siz[i][j][0]; 
				siz[i-1][j][1] %= MOD;
			}
			if(j && b[j] != b[j-1]){
				//dp[i][j][1] += dp[i][j-1][1];
				siz[i][j-1][1] += siz[i][j][1];
				siz[i][j-1][1] %= MOD; 
			}
			if(j && b[j] != a[i]){
				//dp[i][j][1] += dp[i][j-1][0];
				siz[i][j-1][0] += siz[i][j][1];
				siz[i][j-1][0] %= MOD; 
			}
		}
	}
	/*long long res = 0;
	dp[1][0][0] = dp[0][1][1] = 1;
	for(int i = 0;i <= n ;i++){
		for(int j = 0;j <= m ;j++){
			if(i && a[i] != a[i-1]) {
				dp[i][j][0] += dp[i-1][j][0];
				//cout<<dp[i][j][0]<<"\n";
				//dp[i][j][0]表式a串取1到i,b串取1到j,要使merge串的最后一位是a串的最后一位的构造方案数 
				//siz[i-1][j][0] += siz[i][j][0];
				//siz[i-1][j][0] %= MOD;
			} 
			if(i && a[i] != b[j]){
				dp[i][j][0] += dp[i-1][j][1];
				//siz[i-1][j][1] += siz[i][j][0]; 
				//siz[i-1][j][1] %= MOD;
			}
			if(j && b[j] != b[j-1]){
				dp[i][j][1] += dp[i][j-1][1];
				//siz[i][j-1][1] += siz[i][j][1];
				//siz[i][j-1][1] %= MOD; 
			}
			if(j && b[j] != a[i]){
				dp[i][j][1] += dp[i][j-1][0];
				//siz[i][j-1][0] += siz[i][j][1];
				//siz[i][j-1][0] %= MOD; 
			}
			res+= dp[i][j][0] + dp[i][j][1];
		}
	}
	res += 2;*/
	//cout<<"dp[n][m][0] "<<dp[n][m][0]<<" dp[n][m][1] "<<dp[n][m][1]<< " res "<<res<<endl;
	for(int i = 1;i<=n;i++) sza[i] = 1;
	for(int i = n;i;i--){
		if(a[i] != a[i-1]) sza[i-1] += sza[i];
	}
	for(int i = 1;i<=m;i++) szb[i] = 1;
	for(int i = m;i;i--){
		if(b[i] != b[i-1]) szb[i-1] += szb[i];
	}
	for(int l = 1;l <= n;l++){//枚举左右区间 
		for(int r = 1;r <= m;r++){
			//cout<<"l "<<l << " r "<< r << " siz[l][r-1][0] " << siz[l][r-1][0] << " siz[l-1][r][1] "<<siz[l-1][r][1] << 
			//" sza[l] " << sza[l] <<" szb[r] "<< szb[r] << "\n"; 
			ans += siz[l][r-1][0] + siz[l-1][r][1] - sza[l] - szb[r];
			ans = (ans % MOD + MOD) % MOD;
		}
	}
	cout<<ans<<"\n";
	return 0;
}
### Prompt

Your challenge is to write a CPP solution to the following problem:

You are given two strings x and y, both consist only of lowercase Latin letters. Let |s| be the length of string s.

Let's call a sequence a a merging sequence if it consists of exactly |x| zeros and exactly |y| ones in some order.

A merge z is produced from a sequence a by the following rules: 

  * if a_i=0, then remove a letter from the beginning of x and append it to the end of z; 
  * if a_i=1, then remove a letter from the beginning of y and append it to the end of z. 



Two merging sequences a and b are different if there is some position i such that a_i ≠ b_i.

Let's call a string z chaotic if for all i from 2 to |z| z_{i-1} ≠ z_i.

Let s[l,r] for some 1 ≤ l ≤ r ≤ |s| be a substring of consecutive letters of s, starting from position l and ending at position r inclusive.

Let f(l_1, r_1, l_2, r_2) be the number of different merging sequences of x[l_1,r_1] and y[l_2,r_2] that produce chaotic merges. Note that only non-empty substrings of x and y are considered.

Calculate ∑ _{1 ≤ l_1 ≤ r_1 ≤ |x| \\\ 1 ≤ l_2 ≤ r_2 ≤ |y|} f(l_1, r_1, l_2, r_2). Output the answer modulo 998 244 353.

Input

The first line contains a string x (1 ≤ |x| ≤ 1000).

The second line contains a string y (1 ≤ |y| ≤ 1000).

Both strings consist only of lowercase Latin letters.

Output

Print a single integer — the sum of f(l_1, r_1, l_2, r_2) over 1 ≤ l_1 ≤ r_1 ≤ |x| and 1 ≤ l_2 ≤ r_2 ≤ |y| modulo 998 244 353.

Examples

Input


aaa
bb


Output


24


Input


code
forces


Output


1574


Input


aaaaa
aaa


Output


0


Input


justamassivetesttocheck
howwellyouhandlemodulooperations


Output


667387032

Note

In the first example there are: 

  * 6 pairs of substrings "a" and "b", each with valid merging sequences "01" and "10"; 
  * 3 pairs of substrings "a" and "bb", each with a valid merging sequence "101"; 
  * 4 pairs of substrings "aa" and "b", each with a valid merging sequence "010"; 
  * 2 pairs of substrings "aa" and "bb", each with valid merging sequences "0101" and "1010"; 
  * 2 pairs of substrings "aaa" and "b", each with no valid merging sequences; 
  * 1 pair of substrings "aaa" and "bb" with a valid merging sequence "01010"; 



Thus, the answer is 6 ⋅ 2 + 3 ⋅ 1 + 4 ⋅ 1 + 2 ⋅ 2 + 2 ⋅ 0 + 1 ⋅ 1 = 24.

### Response

```cpp
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;
const int MAXN = 1e3+7;
const long long MOD = 998244353;
int n,m;
long long dp[MAXN][MAXN][2],f[MAXN][MAXN][2];
long long siz[MAXN][MAXN][2];
int sza[MAXN],szb[MAXN];

int main()
{
	string a,b;
	cin >> a >> b;
	n = a.length();m = b.length();
	a = " " + a;
	b = " " + b;
	int u,v;
	long long ans = 0;
	for(int i = 0;i<=n;i++) 
		for(int j = 0;j<=m;j++) 
			siz[i][j][0] = siz[i][j][1] = 1;
	for(int i = n;i>=0 ;i--){
		for(int j = m;j>=0 ;j--){
			if(i && a[i] != a[i-1]) {
				//dp[i][j][0] += dp[i-1][j][0];
				//dp[i][j][0]表式a串取1到i,b串取1到j,要使merge串的最后一位是a串的最后一位的构造方案数 
				siz[i-1][j][0] += siz[i][j][0];
				siz[i-1][j][0] %= MOD;
			} 
			if(i && a[i] != b[j]){
				//dp[i][j][0] += dp[i-1][j][1];
				siz[i-1][j][1] += siz[i][j][0]; 
				siz[i-1][j][1] %= MOD;
			}
			if(j && b[j] != b[j-1]){
				//dp[i][j][1] += dp[i][j-1][1];
				siz[i][j-1][1] += siz[i][j][1];
				siz[i][j-1][1] %= MOD; 
			}
			if(j && b[j] != a[i]){
				//dp[i][j][1] += dp[i][j-1][0];
				siz[i][j-1][0] += siz[i][j][1];
				siz[i][j-1][0] %= MOD; 
			}
		}
	}
	/*long long res = 0;
	dp[1][0][0] = dp[0][1][1] = 1;
	for(int i = 0;i <= n ;i++){
		for(int j = 0;j <= m ;j++){
			if(i && a[i] != a[i-1]) {
				dp[i][j][0] += dp[i-1][j][0];
				//cout<<dp[i][j][0]<<"\n";
				//dp[i][j][0]表式a串取1到i,b串取1到j,要使merge串的最后一位是a串的最后一位的构造方案数 
				//siz[i-1][j][0] += siz[i][j][0];
				//siz[i-1][j][0] %= MOD;
			} 
			if(i && a[i] != b[j]){
				dp[i][j][0] += dp[i-1][j][1];
				//siz[i-1][j][1] += siz[i][j][0]; 
				//siz[i-1][j][1] %= MOD;
			}
			if(j && b[j] != b[j-1]){
				dp[i][j][1] += dp[i][j-1][1];
				//siz[i][j-1][1] += siz[i][j][1];
				//siz[i][j-1][1] %= MOD; 
			}
			if(j && b[j] != a[i]){
				dp[i][j][1] += dp[i][j-1][0];
				//siz[i][j-1][0] += siz[i][j][1];
				//siz[i][j-1][0] %= MOD; 
			}
			res+= dp[i][j][0] + dp[i][j][1];
		}
	}
	res += 2;*/
	//cout<<"dp[n][m][0] "<<dp[n][m][0]<<" dp[n][m][1] "<<dp[n][m][1]<< " res "<<res<<endl;
	for(int i = 1;i<=n;i++) sza[i] = 1;
	for(int i = n;i;i--){
		if(a[i] != a[i-1]) sza[i-1] += sza[i];
	}
	for(int i = 1;i<=m;i++) szb[i] = 1;
	for(int i = m;i;i--){
		if(b[i] != b[i-1]) szb[i-1] += szb[i];
	}
	for(int l = 1;l <= n;l++){//枚举左右区间 
		for(int r = 1;r <= m;r++){
			//cout<<"l "<<l << " r "<< r << " siz[l][r-1][0] " << siz[l][r-1][0] << " siz[l-1][r][1] "<<siz[l-1][r][1] << 
			//" sza[l] " << sza[l] <<" szb[r] "<< szb[r] << "\n"; 
			ans += siz[l][r-1][0] + siz[l-1][r][1] - sza[l] - szb[r];
			ans = (ans % MOD + MOD) % MOD;
		}
	}
	cout<<ans<<"\n";
	return 0;
}
```