#include<bits/stdc++.h>
using namespace std;
#define int long long

const int mod=998244353;

inline int read()
{
	int x=0,f=1;
	char c=getchar();
	while(c<'0' || c>'9')
	{
	    if(c=='-')
	        f=-1;
	    c=getchar();
	}
	while(c>='0' && c<='9')
	{
	    x=x*10+c-'0';
	    c=getchar();
	}
	return x*f;
}

const int N=1000+5;
char a[N],b[N];
int f[N][N][2];

signed main()
{
	scanf("%s",a+1); a[0]='-';
	scanf("%s",b+1); b[0]='+';
	int n=strlen(a+1),m=strlen(b+1);
	for(int i=1;i<=n;i++)
	{
		f[i][0][0]=1;
		if(i>1 && a[i]!=a[i-1]) f[i][0][0]+=f[i-1][0][0], f[i][0][0]%=mod;
	}
	for(int i=1;i<=m;i++)
	{
		f[0][i][1]=1;
		if(i>1 && b[i]!=b[i-1]) f[0][i][1]+=f[0][i-1][1], f[0][i][1]%=mod;
	}
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			for(int k=0;k<=1;k++)
			{
				if(!k)
				{
					if(a[i-1]!=a[i]) f[i][j][k]=(f[i][j][k]+f[i-1][j][0])%mod;
					if(b[j]!=a[i]) f[i][j][k]=(f[i][j][k]+f[i-1][j][1])%mod;
					if(b[j]!=a[i] && i>1) f[i][j][k]=(f[i][j][k]+f[0][j][1])%mod;
				}
				else
				{
					if(b[j-1]!=b[j]) f[i][j][k]=(f[i][j][k]+f[i][j-1][1])%mod;
					if(a[i]!=b[j]) f[i][j][k]=(f[i][j][k]+f[i][j-1][0])%mod;
					if(a[i]!=b[j] && j>1) f[i][j][k]=(f[i][j][k]+f[i][0][0])%mod;
				}
			}
	int ans=0;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
		{
			ans=(ans+f[i][j][0])%mod;
			ans=(ans+f[i][j][1])%mod;
		}
	printf("%lld\n",ans);
	return 0;
}

### Prompt

Generate 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>
using namespace std;
#define int long long

const int mod=998244353;

inline int read()
{
	int x=0,f=1;
	char c=getchar();
	while(c<'0' || c>'9')
	{
	    if(c=='-')
	        f=-1;
	    c=getchar();
	}
	while(c>='0' && c<='9')
	{
	    x=x*10+c-'0';
	    c=getchar();
	}
	return x*f;
}

const int N=1000+5;
char a[N],b[N];
int f[N][N][2];

signed main()
{
	scanf("%s",a+1); a[0]='-';
	scanf("%s",b+1); b[0]='+';
	int n=strlen(a+1),m=strlen(b+1);
	for(int i=1;i<=n;i++)
	{
		f[i][0][0]=1;
		if(i>1 && a[i]!=a[i-1]) f[i][0][0]+=f[i-1][0][0], f[i][0][0]%=mod;
	}
	for(int i=1;i<=m;i++)
	{
		f[0][i][1]=1;
		if(i>1 && b[i]!=b[i-1]) f[0][i][1]+=f[0][i-1][1], f[0][i][1]%=mod;
	}
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			for(int k=0;k<=1;k++)
			{
				if(!k)
				{
					if(a[i-1]!=a[i]) f[i][j][k]=(f[i][j][k]+f[i-1][j][0])%mod;
					if(b[j]!=a[i]) f[i][j][k]=(f[i][j][k]+f[i-1][j][1])%mod;
					if(b[j]!=a[i] && i>1) f[i][j][k]=(f[i][j][k]+f[0][j][1])%mod;
				}
				else
				{
					if(b[j-1]!=b[j]) f[i][j][k]=(f[i][j][k]+f[i][j-1][1])%mod;
					if(a[i]!=b[j]) f[i][j][k]=(f[i][j][k]+f[i][j-1][0])%mod;
					if(a[i]!=b[j] && j>1) f[i][j][k]=(f[i][j][k]+f[i][0][0])%mod;
				}
			}
	int ans=0;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
		{
			ans=(ans+f[i][j][0])%mod;
			ans=(ans+f[i][j][1])%mod;
		}
	printf("%lld\n",ans);
	return 0;
}

```