#include<bits/stdc++.h>
#define x first
#define y second
#define bug(x) cerr<<#x<<'='<<x<<' '
#define debug(x) cerr<<#x<<'='<<x<<'\n'
#define FOR(a,b,c) for(int a=(b),a##_end=(c);a<=a##_end;++a)
#define ROF(a,b,c) for(int a=(b),a##_end=(c);a>=a##_end;--a)
using namespace std;
typedef long long ll;
typedef pair<int,int> PII;
const int INF=0x3f3f3f3f,N=1010,P=998244353;
template<class T>inline bool chkmin(T &x,const T &y){return y<x?x=y,1:0;}
template<class T>inline bool chkmax(T &x,const T &y){return x<y?x=y,1:0;}
template<class T>inline void Add(T &x,const T &y){x=(x+y+P)%P;}
int n,m;
char a[N],b[N];
ll dp[N][N][2],ans,res1,res2;
int main(){
	scanf("%s%s",a+1,b+1);
	n=strlen(a+1);
	m=strlen(b+1);
	dp[n+1][m+1][0]=dp[n+1][m+1][1]=1;
	ROF(i,n+1,1)ROF(j,m+1,1)FOR(k,0,1){
		if((k==0&&i==1)||(k==1&&j==1))continue;
		int las=k==0?a[i-1]:b[j-1];
		dp[i][j][k]=1;
		if(i<=n&&las!=a[i])
			Add(dp[i][j][k],dp[i+1][j][0]);
		if(j<=m&&las!=b[j])
			Add(dp[i][j][k],dp[i][j+1][1]);
	}
	dp[1][1][0]=(dp[2][1][0]+dp[1][2][1])%P;
	FOR(i,1,n)FOR(j,1,m){
		Add(ans,dp[i+1][j][0]);
		Add(ans,dp[i][j+1][1]);
	}
	FOR(i,1,n){
		int j=i+1;
		for(;j<=n;++j)
			if(a[j-1]==a[j])
				break;
		res1+=j-i;
	}
	FOR(i,1,m){
		int j=i+1;
		for(;j<=m;++j)
			if(b[j-1]==b[j])
				break;
		res2+=j-i;
	}
	ans=ans-n*res2-m*res1;
	printf("%lld\n",(ans%P+P)%P);
	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<bits/stdc++.h>
#define x first
#define y second
#define bug(x) cerr<<#x<<'='<<x<<' '
#define debug(x) cerr<<#x<<'='<<x<<'\n'
#define FOR(a,b,c) for(int a=(b),a##_end=(c);a<=a##_end;++a)
#define ROF(a,b,c) for(int a=(b),a##_end=(c);a>=a##_end;--a)
using namespace std;
typedef long long ll;
typedef pair<int,int> PII;
const int INF=0x3f3f3f3f,N=1010,P=998244353;
template<class T>inline bool chkmin(T &x,const T &y){return y<x?x=y,1:0;}
template<class T>inline bool chkmax(T &x,const T &y){return x<y?x=y,1:0;}
template<class T>inline void Add(T &x,const T &y){x=(x+y+P)%P;}
int n,m;
char a[N],b[N];
ll dp[N][N][2],ans,res1,res2;
int main(){
	scanf("%s%s",a+1,b+1);
	n=strlen(a+1);
	m=strlen(b+1);
	dp[n+1][m+1][0]=dp[n+1][m+1][1]=1;
	ROF(i,n+1,1)ROF(j,m+1,1)FOR(k,0,1){
		if((k==0&&i==1)||(k==1&&j==1))continue;
		int las=k==0?a[i-1]:b[j-1];
		dp[i][j][k]=1;
		if(i<=n&&las!=a[i])
			Add(dp[i][j][k],dp[i+1][j][0]);
		if(j<=m&&las!=b[j])
			Add(dp[i][j][k],dp[i][j+1][1]);
	}
	dp[1][1][0]=(dp[2][1][0]+dp[1][2][1])%P;
	FOR(i,1,n)FOR(j,1,m){
		Add(ans,dp[i+1][j][0]);
		Add(ans,dp[i][j+1][1]);
	}
	FOR(i,1,n){
		int j=i+1;
		for(;j<=n;++j)
			if(a[j-1]==a[j])
				break;
		res1+=j-i;
	}
	FOR(i,1,m){
		int j=i+1;
		for(;j<=m;++j)
			if(b[j-1]==b[j])
				break;
		res2+=j-i;
	}
	ans=ans-n*res2-m*res1;
	printf("%lld\n",(ans%P+P)%P);
	return 0;
}

```