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

const int N = 1005;
const int p = 998244353;
int n, l1, l2, ans, a[N], b[N], f[N][N][2];
char s1[N], s2[N];

int main()
{
	scanf("%s", s1), l1 = strlen(s1);
	scanf("%s", s2), l2 = strlen(s2);
	for (Re i = 0; i < l1; ++i)
	{
		if (!i) a[i] = 1;
		else if (s1[i] ^ s1[i - 1]) a[i] = a[i - 1] + 1;
		else a[i] = 1;
	}
	for (Re i = 0; i < l2; ++i)
	{
		if (!i) b[i] = 1;
		else if (s2[i] ^ s2[i - 1]) b[i] = b[i - 1] + 1;
		else b[i] = 1;
	}
	for (Re i = 0; i < l1; ++i)
		for (Re j = 0; j < l2; ++j)
		{
			if (s1[i] ^ s2[j]) f[i][j][0] = b[j], f[i][j][1] = a[i];
			if (i)
			{
				if (s1[i] ^ s1[i - 1])
				{
					f[i][j][0] += f[i - 1][j][0];
					if (f[i][j][0] >= p) f[i][j][0] -= p;
				}
				if (s1[i] ^ s2[j])
				{
					f[i][j][0] += f[i - 1][j][1];
					if (f[i][j][0] >= p) f[i][j][0] -= p;
				}
			}
			if (j)
			{
				if (s2[j] ^ s2[j - 1])
				{
					f[i][j][1] += f[i][j - 1][1];
					if (f[i][j][1] >= p) f[i][j][1] -= p;
				}
				if (s1[i] ^ s2[j])
				{
					f[i][j][1] += f[i][j - 1][0];
					if (f[i][j][1] >= p) f[i][j][1] -= p;
				}
			}
			ans += f[i][j][0];
			if (ans >= p) ans -= p;
			ans += f[i][j][1];
			if (ans >= p) ans -= p;
		}
	printf("%d", 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<bits/stdc++.h>
using namespace std;
#define Re register int
typedef long long ll;

const int N = 1005;
const int p = 998244353;
int n, l1, l2, ans, a[N], b[N], f[N][N][2];
char s1[N], s2[N];

int main()
{
	scanf("%s", s1), l1 = strlen(s1);
	scanf("%s", s2), l2 = strlen(s2);
	for (Re i = 0; i < l1; ++i)
	{
		if (!i) a[i] = 1;
		else if (s1[i] ^ s1[i - 1]) a[i] = a[i - 1] + 1;
		else a[i] = 1;
	}
	for (Re i = 0; i < l2; ++i)
	{
		if (!i) b[i] = 1;
		else if (s2[i] ^ s2[i - 1]) b[i] = b[i - 1] + 1;
		else b[i] = 1;
	}
	for (Re i = 0; i < l1; ++i)
		for (Re j = 0; j < l2; ++j)
		{
			if (s1[i] ^ s2[j]) f[i][j][0] = b[j], f[i][j][1] = a[i];
			if (i)
			{
				if (s1[i] ^ s1[i - 1])
				{
					f[i][j][0] += f[i - 1][j][0];
					if (f[i][j][0] >= p) f[i][j][0] -= p;
				}
				if (s1[i] ^ s2[j])
				{
					f[i][j][0] += f[i - 1][j][1];
					if (f[i][j][0] >= p) f[i][j][0] -= p;
				}
			}
			if (j)
			{
				if (s2[j] ^ s2[j - 1])
				{
					f[i][j][1] += f[i][j - 1][1];
					if (f[i][j][1] >= p) f[i][j][1] -= p;
				}
				if (s1[i] ^ s2[j])
				{
					f[i][j][1] += f[i][j - 1][0];
					if (f[i][j][1] >= p) f[i][j][1] -= p;
				}
			}
			ans += f[i][j][0];
			if (ans >= p) ans -= p;
			ans += f[i][j][1];
			if (ans >= p) ans -= p;
		}
	printf("%d", ans);
	return 0;
}
```