#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<cctype>
#include<cstdlib>
#include<vector>
#include<algorithm>
#include<queue>
#define fs(i,x,y) for(int i=(x),_=(y);i<=_;++i)
#define fn(i,x,y) for(int i=(x),_=(y);i>=_;--i)
#define tor(i,v,x) for(int i=head[x],v=to[i];i;i=nxt[i],v=to[i])
#define ls(x) ch[x][0]
#define rs(x) ch[x][1]
#define mpi(x,y) make_pair(x,y)
#define pi pair<int,int>
//#define int ll
#define DEBUG printf("%s %d\n",__FUNCTION__,__LINE__)
using namespace std;
typedef long long ll;
typedef double db;
template<typename T>
void read(T &x){
	x=0;char ch=getchar();bool f=0;
	while(ch<'0'||ch>'9'){
		if(ch=='-')f=1;
		ch=getchar();
	}
	while(ch>='0'&&ch<='9')x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
	x=f?-x:x;
}

const int N=1e3+7,P=998244353;
int n,m;
char s[N],t[N];
int f[N][N][2][4];

void add(int &x,int y){x+=y;if(x>=P)x-=P;}

signed main(){
	scanf("%s",s+1);scanf("%s",t+1);
	n=strlen(s+1);m=strlen(t+1);
	fs(i,0,n)fs(j,0,m){
		if(i<n)add(f[i+1][j][0][1],1);
		if(j<m)add(f[i][j+1][1][2],1);//需要标记最后取哪个
		//而且都取了的方案才是合法
		fs(st,0,3){
			if(i<n&&s[i]!=s[i+1])add(f[i+1][j][0][st|1],f[i][j][0][st]);
			if(j<m&&s[i]!=t[j+1])add(f[i][j+1][1][st|2],f[i][j][0][st]);
			if(i<n&&t[j]!=s[i+1])add(f[i+1][j][0][st|1],f[i][j][1][st]);
			if(j<m&&t[j]!=t[j+1])add(f[i][j+1][1][st|2],f[i][j][1][st]);
		}
	}
	int ans=0;
	fs(i,1,n)fs(j,1,m)add(ans,f[i][j][0][3]),add(ans,f[i][j][1][3]);
	printf("%d\n",ans);
	return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<cctype>
#include<cstdlib>
#include<vector>
#include<algorithm>
#include<queue>
#define fs(i,x,y) for(int i=(x),_=(y);i<=_;++i)
#define fn(i,x,y) for(int i=(x),_=(y);i>=_;--i)
#define tor(i,v,x) for(int i=head[x],v=to[i];i;i=nxt[i],v=to[i])
#define ls(x) ch[x][0]
#define rs(x) ch[x][1]
#define mpi(x,y) make_pair(x,y)
#define pi pair<int,int>
//#define int ll
#define DEBUG printf("%s %d\n",__FUNCTION__,__LINE__)
using namespace std;
typedef long long ll;
typedef double db;
template<typename T>
void read(T &x){
	x=0;char ch=getchar();bool f=0;
	while(ch<'0'||ch>'9'){
		if(ch=='-')f=1;
		ch=getchar();
	}
	while(ch>='0'&&ch<='9')x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
	x=f?-x:x;
}

const int N=1e3+7,P=998244353;
int n,m;
char s[N],t[N];
int f[N][N][2][4];

void add(int &x,int y){x+=y;if(x>=P)x-=P;}

signed main(){
	scanf("%s",s+1);scanf("%s",t+1);
	n=strlen(s+1);m=strlen(t+1);
	fs(i,0,n)fs(j,0,m){
		if(i<n)add(f[i+1][j][0][1],1);
		if(j<m)add(f[i][j+1][1][2],1);//需要标记最后取哪个
		//而且都取了的方案才是合法
		fs(st,0,3){
			if(i<n&&s[i]!=s[i+1])add(f[i+1][j][0][st|1],f[i][j][0][st]);
			if(j<m&&s[i]!=t[j+1])add(f[i][j+1][1][st|2],f[i][j][0][st]);
			if(i<n&&t[j]!=s[i+1])add(f[i+1][j][0][st|1],f[i][j][1][st]);
			if(j<m&&t[j]!=t[j+1])add(f[i][j+1][1][st|2],f[i][j][1][st]);
		}
	}
	int ans=0;
	fs(i,1,n)fs(j,1,m)add(ans,f[i][j][0][3]),add(ans,f[i][j][1][3]);
	printf("%d\n",ans);
	return 0;
}
```