#include<bits/stdc++.h>
#define ll long long
const int N = 1010;
const ll mod = 998244353;
char x[N],y[N];
int n,m;
ll dp[N][N][2],ans;
int main(){
	scanf(" %s",x+1);
	scanf(" %s",y+1);
	n = strlen(x+1);
	m = strlen(y+1);
	for(int i = 0 ; i <= n ; ++i){
		for(int j = 0 ; j <= m ;++j){
			if(i>=1) dp[i][j][0]+=1;
			if(j>=1) dp[i][j][1]+=1;
			if(i-1>=0&&x[i]!=x[i-1])
				dp[i][j][0]+=dp[i-1][j][0];
			if(i-1>=0&&x[i]!=y[j])
				dp[i][j][0]+=dp[i-1][j][1];
			if(j-1>=0&&y[j]!=x[i])
				dp[i][j][1]+=dp[i][j-1][0];
			if(j-1>=0&&y[j]!=y[j-1])
				dp[i][j][1]+=dp[i][j-1][1];
			dp[i][j][0]%=mod;
			dp[i][j][1]%=mod;
			ans = (ans + dp[i][j][0])%mod;
			ans = (ans + dp[i][j][1])%mod;
		}
	}
	//printf("debug : %lld %lld %lld %lld\n",dp[1][0][0],dp[0][1][1],dp[1][1][0],dp[1][1][1]);
	for(int i = 1 ; i <= n ; ++i){
		int j = i ;
		while(j+1<=n&&x[j+1]!=x[j]) ++j ;
		ll len = j - i + 1;
		ll val = (len)*(len+1)/2;
		val%=mod;
		val = (val * (m+1)) %mod ;
		//printf("x:val %d\n",val);
		ans = (ans - val + mod)%mod;
		i = j ; 
	}
	for(int i = 1 ; i <= m ; ++i){
		int j = i ;
		while(j+1<=m&&y[j+1]!=y[j]) ++j ;
		ll len = j - i + 1;
		ll val = (len)*(len+1)/2;
		val%=mod;
		val = (val * (n+1)) %mod ;
		//printf("y:val %d\n",val);
		ans = (ans - val + mod)%mod;
		i = j ; 
	}
	printf("%lld",ans);
	return 0;
} 
### Prompt

Develop a solution in CPP to the problem described below:

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<bits/stdc++.h>
#define ll long long
const int N = 1010;
const ll mod = 998244353;
char x[N],y[N];
int n,m;
ll dp[N][N][2],ans;
int main(){
	scanf(" %s",x+1);
	scanf(" %s",y+1);
	n = strlen(x+1);
	m = strlen(y+1);
	for(int i = 0 ; i <= n ; ++i){
		for(int j = 0 ; j <= m ;++j){
			if(i>=1) dp[i][j][0]+=1;
			if(j>=1) dp[i][j][1]+=1;
			if(i-1>=0&&x[i]!=x[i-1])
				dp[i][j][0]+=dp[i-1][j][0];
			if(i-1>=0&&x[i]!=y[j])
				dp[i][j][0]+=dp[i-1][j][1];
			if(j-1>=0&&y[j]!=x[i])
				dp[i][j][1]+=dp[i][j-1][0];
			if(j-1>=0&&y[j]!=y[j-1])
				dp[i][j][1]+=dp[i][j-1][1];
			dp[i][j][0]%=mod;
			dp[i][j][1]%=mod;
			ans = (ans + dp[i][j][0])%mod;
			ans = (ans + dp[i][j][1])%mod;
		}
	}
	//printf("debug : %lld %lld %lld %lld\n",dp[1][0][0],dp[0][1][1],dp[1][1][0],dp[1][1][1]);
	for(int i = 1 ; i <= n ; ++i){
		int j = i ;
		while(j+1<=n&&x[j+1]!=x[j]) ++j ;
		ll len = j - i + 1;
		ll val = (len)*(len+1)/2;
		val%=mod;
		val = (val * (m+1)) %mod ;
		//printf("x:val %d\n",val);
		ans = (ans - val + mod)%mod;
		i = j ; 
	}
	for(int i = 1 ; i <= m ; ++i){
		int j = i ;
		while(j+1<=m&&y[j+1]!=y[j]) ++j ;
		ll len = j - i + 1;
		ll val = (len)*(len+1)/2;
		val%=mod;
		val = (val * (n+1)) %mod ;
		//printf("y:val %d\n",val);
		ans = (ans - val + mod)%mod;
		i = j ; 
	}
	printf("%lld",ans);
	return 0;
} 
```