#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
int read()
{
	int res = 0, f = 0; char c = getchar();
	while(!isdigit(c) && c != '-') c = getchar();
	if(c == '-') f++, c = getchar();
	while(isdigit(c)) res = (res<<3)+(res<<1)+(c^48), c = getchar();
	return f ? -res : res;
} 

const int maxn = 1e3+10;
const int mod = 998244353;

int n, m, dpa[maxn][maxn], dpb[maxn][maxn];
int pra[maxn], prb[maxn];
char a[maxn], b[maxn];

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

int main()
{
	scanf("%s%s", a+1, b+1);
	n = strlen(a+1);
	m = strlen(b+1);
	for(int i = 1; i <= n; i++) 
	{
		pra[i] = a[i] == a[i-1] ? i-1 : pra[i-1];
	}
	for(int i = 1; i <= m; i++) 
	{
		prb[i] = b[i] == b[i-1] ? i-1 : prb[i-1];
	}
	int ans = 0;
	for(int i = 1; i <= n; i++)
	for(int j = 1; j <= m; j++)
	{
		if(a[i] != b[j]) 
		{
			add(dpa[i][j], j-prb[j]);
			add(dpb[i][j], i-pra[i]);
		}
		add(ans, dpa[i][j]);
		add(ans, dpb[i][j]);
		if(i < n && a[i] != a[i+1]) add(dpa[i+1][j], dpa[i][j]);
		if(j < m && a[i] != b[j+1]) add(dpb[i][j+1], dpa[i][j]);
		if(i < n && b[j] != a[i+1]) add(dpa[i+1][j], dpb[i][j]);
		if(j < m && b[j] != b[j+1]) add(dpb[i][j+1], dpb[i][j]);
	}
	printf("%d\n", ans);
	return 0;
}
/*
aba
bb

*/

### 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 <bits/stdc++.h>
using namespace std;
typedef long long LL;
int read()
{
	int res = 0, f = 0; char c = getchar();
	while(!isdigit(c) && c != '-') c = getchar();
	if(c == '-') f++, c = getchar();
	while(isdigit(c)) res = (res<<3)+(res<<1)+(c^48), c = getchar();
	return f ? -res : res;
} 

const int maxn = 1e3+10;
const int mod = 998244353;

int n, m, dpa[maxn][maxn], dpb[maxn][maxn];
int pra[maxn], prb[maxn];
char a[maxn], b[maxn];

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

int main()
{
	scanf("%s%s", a+1, b+1);
	n = strlen(a+1);
	m = strlen(b+1);
	for(int i = 1; i <= n; i++) 
	{
		pra[i] = a[i] == a[i-1] ? i-1 : pra[i-1];
	}
	for(int i = 1; i <= m; i++) 
	{
		prb[i] = b[i] == b[i-1] ? i-1 : prb[i-1];
	}
	int ans = 0;
	for(int i = 1; i <= n; i++)
	for(int j = 1; j <= m; j++)
	{
		if(a[i] != b[j]) 
		{
			add(dpa[i][j], j-prb[j]);
			add(dpb[i][j], i-pra[i]);
		}
		add(ans, dpa[i][j]);
		add(ans, dpb[i][j]);
		if(i < n && a[i] != a[i+1]) add(dpa[i+1][j], dpa[i][j]);
		if(j < m && a[i] != b[j+1]) add(dpb[i][j+1], dpa[i][j]);
		if(i < n && b[j] != a[i+1]) add(dpa[i+1][j], dpb[i][j]);
		if(j < m && b[j] != b[j+1]) add(dpb[i][j+1], dpb[i][j]);
	}
	printf("%d\n", ans);
	return 0;
}
/*
aba
bb

*/

```