#include<bits/stdc++.h>
using namespace std;
#define M 500000
typedef long long ll;
typedef unsigned long long ull;
inline ll read() {
	char ch;
	ll f = 1, res = 0;
	while ((ch = getchar()) < '0' || ch > '9')
		if (ch == '-') f = -1;
	while (ch >= '0' && ch <= '9') {
		res = (res << 1) + (res << 3) + (ch ^ 48);
		ch = getchar();
	}
	return f * res;
}
inline void pt(ll u,ll mod) {
	printf("%lld", u);
	if (mod == 1)
		putchar(' ');
	else if (mod == 2)
		putchar('\n');
}
char s1[1010];
char s2[1010];
ll dp[1010][1010][2][2][2];
ll mod=998244353;
//(x|y)(0|1)(have x or not )(0|1),(have y or not)(0|1)
int main() {
	//freopen("../in", "r", stdin);
	//freopen("../out", "w", stdout);
	scanf("%s%s", s1 + 1, s2 + 1);
	ll lens1 = strlen(s1 + 1);
	ll lens2 = strlen(s2 + 1);

	ll ans = 0;
	for (ll i = 1; i <= lens1; i++) {
		if (s1[i] != s1[i - 1])
			dp[i][0][0][1][0] += dp[i - 1][0][0][1][0];
		dp[i][0][0][1][0]++;
	}

	for (ll i = 1; i <= lens2; i++) {
		if (s2[i] != s2[i - 1])
			dp[0][i][1][0][1] += dp[0][i - 1][1][0][1];
		dp[0][i][1][0][1]++;
	}

	for (ll i = 1; i <= lens1; i++) {
		for (ll j = 1; j <= lens2; j++) {

			if(s1[i] != s2[j]) {
				dp[i][j][0][1][1] += dp[i - 1][j][1][0][1] + dp[i - 1][j][1][1][1];
				dp[i][j][1][1][1] += dp[i][j - 1][0][1][0] + dp[i][j - 1][0][1][1];
			}

			if (s1[i] != s1[i - 1]) {
				dp[i][j][0][1][0] += dp[i - 1][j][0][1][0];
				dp[i][j][0][1][1] += dp[i - 1][j][0][1][1];
			}

			if (s2[j] != s2[j - 1]) {
				dp[i][j][1][0][1] += dp[i][j - 1][1][0][1];
				dp[i][j][1][1][1] += dp[i][j - 1][1][1][1];
			}

			dp[i][j][0][1][0]++;
			dp[i][j][1][0][1]++;

			dp[i][j][0][1][0] %= mod;
			dp[i][j][1][0][1] %= mod;
			dp[i][j][0][1][1] %= mod;
			dp[i][j][1][1][1] %= mod;

			ans += dp[i][j][0][1][1] + dp[i][j][1][1][1];
			ans %= mod;
//			cout << i << ':' << j << '=' << dp[i][j][0][1][1] + dp[i][j][1][1][1] << endl;
		}
	}
	pt(ans, 2);
	return 0;
}
### Prompt

Your task is to create 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 M 500000
typedef long long ll;
typedef unsigned long long ull;
inline ll read() {
	char ch;
	ll f = 1, res = 0;
	while ((ch = getchar()) < '0' || ch > '9')
		if (ch == '-') f = -1;
	while (ch >= '0' && ch <= '9') {
		res = (res << 1) + (res << 3) + (ch ^ 48);
		ch = getchar();
	}
	return f * res;
}
inline void pt(ll u,ll mod) {
	printf("%lld", u);
	if (mod == 1)
		putchar(' ');
	else if (mod == 2)
		putchar('\n');
}
char s1[1010];
char s2[1010];
ll dp[1010][1010][2][2][2];
ll mod=998244353;
//(x|y)(0|1)(have x or not )(0|1),(have y or not)(0|1)
int main() {
	//freopen("../in", "r", stdin);
	//freopen("../out", "w", stdout);
	scanf("%s%s", s1 + 1, s2 + 1);
	ll lens1 = strlen(s1 + 1);
	ll lens2 = strlen(s2 + 1);

	ll ans = 0;
	for (ll i = 1; i <= lens1; i++) {
		if (s1[i] != s1[i - 1])
			dp[i][0][0][1][0] += dp[i - 1][0][0][1][0];
		dp[i][0][0][1][0]++;
	}

	for (ll i = 1; i <= lens2; i++) {
		if (s2[i] != s2[i - 1])
			dp[0][i][1][0][1] += dp[0][i - 1][1][0][1];
		dp[0][i][1][0][1]++;
	}

	for (ll i = 1; i <= lens1; i++) {
		for (ll j = 1; j <= lens2; j++) {

			if(s1[i] != s2[j]) {
				dp[i][j][0][1][1] += dp[i - 1][j][1][0][1] + dp[i - 1][j][1][1][1];
				dp[i][j][1][1][1] += dp[i][j - 1][0][1][0] + dp[i][j - 1][0][1][1];
			}

			if (s1[i] != s1[i - 1]) {
				dp[i][j][0][1][0] += dp[i - 1][j][0][1][0];
				dp[i][j][0][1][1] += dp[i - 1][j][0][1][1];
			}

			if (s2[j] != s2[j - 1]) {
				dp[i][j][1][0][1] += dp[i][j - 1][1][0][1];
				dp[i][j][1][1][1] += dp[i][j - 1][1][1][1];
			}

			dp[i][j][0][1][0]++;
			dp[i][j][1][0][1]++;

			dp[i][j][0][1][0] %= mod;
			dp[i][j][1][0][1] %= mod;
			dp[i][j][0][1][1] %= mod;
			dp[i][j][1][1][1] %= mod;

			ans += dp[i][j][0][1][1] + dp[i][j][1][1][1];
			ans %= mod;
//			cout << i << ':' << j << '=' << dp[i][j][0][1][1] + dp[i][j][1][1][1] << endl;
		}
	}
	pt(ans, 2);
	return 0;
}
```