#include <cstdio>
#include <cstring>
#include <cctype>
#include <algorithm>
#include <vector>
#include <set>
using namespace std;

namespace thecold
{
	typedef long long ll;

	inline ll read()
	{
		ll res = 0; bool flag = false;
		char c = getchar();
		while(!isdigit(c))
			flag = (c == '-') , c = getchar();
		while(isdigit(c))
			res = res * 10 + (c ^ 48) , c = getchar();
		return !flag ? res : -res;
	}

	const int Max_n = 1e3 + 5;
	const int jzp = 998244353;

	int n , m , f[Max_n][Max_n][2];

	inline void MAX(int &x , const int y)
	{
		if(x < y) x = y;
	}

	inline void MIN(int &x , const int y)
	{
		if(x > y) x = y;
	}
	
	inline void ADD(int &x , const int y)
	{
		x += y; if(x >= jzp) x -= jzp;
	}
	
	inline int add(const int x , const int y)
	{
		int ret = x + y;
		if(ret >= jzp) ret -= jzp;
		return ret;
	}
	
	char a[Max_n] , b[Max_n];

	inline void main()
	{
		scanf("%s" , a + 1); scanf("%s" , b + 1);
		n = strlen(a + 1) , m = strlen(b + 1);
		for(int i = 1 ; i <= n ; ++ i)
			for(int j = 1 ; j <= m ; ++ j)
				f[i][j - 1][0] = f[i - 1][j][1] = 1;
		for(int i = 0 ; i <= n ; ++ i)
			for(int j = 0 ; j <= m ; ++ j)
			{
				if(i > 0 && a[i] != a[i - 1])
					ADD(f[i][j][0] , f[i - 1][j][0]);
				if(i > 0 && a[i] != b[j])
					ADD(f[i][j][0] , f[i - 1][j][1]);
				if(j > 0 && b[j] != b[j - 1])
					ADD(f[i][j][1] , f[i][j - 1][1]);
				if(j > 0 && b[j] != a[i])
					ADD(f[i][j][1] , f[i][j - 1][0]);
			}
		int cnt = 0;
		for(int i = 1 ; i <= n ; ++ i)
		{
			int ct = 1;
			for(int j = i + 1 ; j <= n ; ++ j)
				if(a[j] != a[j - 1]) ++ ct;
				else break;
			ADD(cnt , 1ll * ct * m % jzp);
		}
		for(int i = 1 ; i <= m ; ++ i)
		{
			int ct = 1;
			for(int j = i + 1 ; j <= m ; ++ j)
				if(b[j] != b[j - 1]) ++ ct;
				else break;
			ADD(cnt , 1ll * ct * n % jzp);
		}
		int ans = 0;
		for(int i = 0 ; i <= n ; ++ i)
			for(int j = 0 ; j <= m ; ++ j)
				ADD(ans , add(f[i][j][0] , f[i][j][1]));
		printf("%d\n" , add(ans , jzp - cnt));
	}
}

int main()
{
	thecold::main();
	return 0;
}

### Prompt

Create a solution in CPP for 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 <cstdio>
#include <cstring>
#include <cctype>
#include <algorithm>
#include <vector>
#include <set>
using namespace std;

namespace thecold
{
	typedef long long ll;

	inline ll read()
	{
		ll res = 0; bool flag = false;
		char c = getchar();
		while(!isdigit(c))
			flag = (c == '-') , c = getchar();
		while(isdigit(c))
			res = res * 10 + (c ^ 48) , c = getchar();
		return !flag ? res : -res;
	}

	const int Max_n = 1e3 + 5;
	const int jzp = 998244353;

	int n , m , f[Max_n][Max_n][2];

	inline void MAX(int &x , const int y)
	{
		if(x < y) x = y;
	}

	inline void MIN(int &x , const int y)
	{
		if(x > y) x = y;
	}
	
	inline void ADD(int &x , const int y)
	{
		x += y; if(x >= jzp) x -= jzp;
	}
	
	inline int add(const int x , const int y)
	{
		int ret = x + y;
		if(ret >= jzp) ret -= jzp;
		return ret;
	}
	
	char a[Max_n] , b[Max_n];

	inline void main()
	{
		scanf("%s" , a + 1); scanf("%s" , b + 1);
		n = strlen(a + 1) , m = strlen(b + 1);
		for(int i = 1 ; i <= n ; ++ i)
			for(int j = 1 ; j <= m ; ++ j)
				f[i][j - 1][0] = f[i - 1][j][1] = 1;
		for(int i = 0 ; i <= n ; ++ i)
			for(int j = 0 ; j <= m ; ++ j)
			{
				if(i > 0 && a[i] != a[i - 1])
					ADD(f[i][j][0] , f[i - 1][j][0]);
				if(i > 0 && a[i] != b[j])
					ADD(f[i][j][0] , f[i - 1][j][1]);
				if(j > 0 && b[j] != b[j - 1])
					ADD(f[i][j][1] , f[i][j - 1][1]);
				if(j > 0 && b[j] != a[i])
					ADD(f[i][j][1] , f[i][j - 1][0]);
			}
		int cnt = 0;
		for(int i = 1 ; i <= n ; ++ i)
		{
			int ct = 1;
			for(int j = i + 1 ; j <= n ; ++ j)
				if(a[j] != a[j - 1]) ++ ct;
				else break;
			ADD(cnt , 1ll * ct * m % jzp);
		}
		for(int i = 1 ; i <= m ; ++ i)
		{
			int ct = 1;
			for(int j = i + 1 ; j <= m ; ++ j)
				if(b[j] != b[j - 1]) ++ ct;
				else break;
			ADD(cnt , 1ll * ct * n % jzp);
		}
		int ans = 0;
		for(int i = 0 ; i <= n ; ++ i)
			for(int j = 0 ; j <= m ; ++ j)
				ADD(ans , add(f[i][j][0] , f[i][j][1]));
		printf("%d\n" , add(ans , jzp - cnt));
	}
}

int main()
{
	thecold::main();
	return 0;
}

```