#include "bits/stdc++.h"
#define MAXN 100009
#define INF 998244353
#define mp(x,y) make_pair(x,y)
#define all(v) v.begin(),v.end()
#define pb(x) push_back(x)
#define wr cout<<"----------------"<<endl;
#define ppb() pop_back()
#define tr(ii,c) for(__typeof((c).begin()) ii=(c).begin();ii!=(c).end();ii++)
#define ff first
#define ss second
#define my_little_dodge 46
#define debug(x)  cerr<< #x <<" = "<< x<<endl;
using namespace std;

typedef long long ll;
typedef pair<int,int> PII;
template<class T>bool umin(T& a,T b){if(a>b){a=b;return 1;}return 0;}
template<class T>bool umax(T& a,T b){if(a<b){a=b;return 1;}return 0;}
int mod(ll x){x%=INF;
	if(x<0)x+=INF;
	return x;	
}
const int N=1004;
char s[N],t[N];
int dp[N][N][26],n,m;
int rec(int x,int y,int last){
	if(x==n and y==m)return 1;
	int &ret=dp[x][y][last];
	if(~ret)return ret;ret=1;
	if(x<n and s[x]-'a'!=last)ret=mod(ret+rec(x+1,y,s[x]-'a'));
	if(y<m and t[y]-'a'!=last)ret=mod(ret+rec(x,y+1,t[y]-'a'));	
	return ret;
}
int main(){
    //freopen("file.in", "r", stdin);
    scanf("%s",s);n=strlen(s);
    scanf("%s",t);m=strlen(t);
    memset(dp,-1,sizeof dp);
    int ans=0;
    for(int i=0;i<n;i++)
    	for(int j=0;j<m;j++){
    		ans=mod(ans+rec(i+1,j,s[i]-'a'));
    		ans=mod(ans+rec(i,j+1,t[j]-'a'));
    	}
    for(int i=0;i<n;i++){
    	int l=i+1;	
    	while(l<n and s[l]!=s[l-1])l++;
    	ans=(ans-(l-i)*1LL*m);
    }
	for(int j=0;j<m;j++){
		int r=j+1;
		while(r<m and t[r]!=t[r-1])r++;
		ans=(ans-(r-j)*1LL*n);
	}	
    printf("%d\n",ans);
	return 0;
}

### Prompt

Please formulate 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 MAXN 100009
#define INF 998244353
#define mp(x,y) make_pair(x,y)
#define all(v) v.begin(),v.end()
#define pb(x) push_back(x)
#define wr cout<<"----------------"<<endl;
#define ppb() pop_back()
#define tr(ii,c) for(__typeof((c).begin()) ii=(c).begin();ii!=(c).end();ii++)
#define ff first
#define ss second
#define my_little_dodge 46
#define debug(x)  cerr<< #x <<" = "<< x<<endl;
using namespace std;

typedef long long ll;
typedef pair<int,int> PII;
template<class T>bool umin(T& a,T b){if(a>b){a=b;return 1;}return 0;}
template<class T>bool umax(T& a,T b){if(a<b){a=b;return 1;}return 0;}
int mod(ll x){x%=INF;
	if(x<0)x+=INF;
	return x;	
}
const int N=1004;
char s[N],t[N];
int dp[N][N][26],n,m;
int rec(int x,int y,int last){
	if(x==n and y==m)return 1;
	int &ret=dp[x][y][last];
	if(~ret)return ret;ret=1;
	if(x<n and s[x]-'a'!=last)ret=mod(ret+rec(x+1,y,s[x]-'a'));
	if(y<m and t[y]-'a'!=last)ret=mod(ret+rec(x,y+1,t[y]-'a'));	
	return ret;
}
int main(){
    //freopen("file.in", "r", stdin);
    scanf("%s",s);n=strlen(s);
    scanf("%s",t);m=strlen(t);
    memset(dp,-1,sizeof dp);
    int ans=0;
    for(int i=0;i<n;i++)
    	for(int j=0;j<m;j++){
    		ans=mod(ans+rec(i+1,j,s[i]-'a'));
    		ans=mod(ans+rec(i,j+1,t[j]-'a'));
    	}
    for(int i=0;i<n;i++){
    	int l=i+1;	
    	while(l<n and s[l]!=s[l-1])l++;
    	ans=(ans-(l-i)*1LL*m);
    }
	for(int j=0;j<m;j++){
		int r=j+1;
		while(r<m and t[r]!=t[r-1])r++;
		ans=(ans-(r-j)*1LL*n);
	}	
    printf("%d\n",ans);
	return 0;
}

```