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

const int mod = 998244353;
const int Maxn = 1005;
const int Maxl = 26;

string A;
string B;
int dp[Maxn][Maxn][Maxl];
int res;

int main()
{
	cin >> A;
	cin >> B;
	int lst = Maxl, cnt = 0;
	for (int i = 0; i < A.length(); i++) {
		if (lst == A[i] - 'a') cnt = 0;
		lst = A[i] - 'a'; cnt++;
		for (int j = 0; j < B.length(); j++)
			if (lst != B[j] - 'a')
				dp[i + 1][j + 1][B[j] - 'a'] = (dp[i + 1][j + 1][B[j] - 'a'] + cnt) % mod;
	}
	lst = Maxl, cnt = 0;
	for (int j = 0; j < B.length(); j++) {
		if (lst == B[j] - 'a') cnt = 0;
		lst = B[j] - 'a'; cnt++;
		for (int i = 0; i < A.length(); i++)
			if (lst != A[i] - 'a')
				dp[i + 1][j + 1][A[i] - 'a'] = (dp[i + 1][j + 1][A[i] - 'a'] + cnt) % mod;
	}
	for (int i = 0; i <= A.length(); i++)
		for (int j = 0; j <= B.length(); j++)
			for (int l = 0; l < Maxl; l++) if (dp[i][j][l]) {
				res = (res + dp[i][j][l]) % mod;
				if (i < A.length() && A[i] - 'a' != l)
					dp[i + 1][j][A[i] - 'a'] = (dp[i + 1][j][A[i] - 'a'] + dp[i][j][l]) % mod;
				if (j < B.length() && B[j] - 'a' != l)
					dp[i][j + 1][B[j] - 'a'] = (dp[i][j + 1][B[j] - 'a'] + dp[i][j][l]) % mod;
			}
	printf("%d\n", res);
    return 0;
}

### Prompt

Please formulate 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;

const int mod = 998244353;
const int Maxn = 1005;
const int Maxl = 26;

string A;
string B;
int dp[Maxn][Maxn][Maxl];
int res;

int main()
{
	cin >> A;
	cin >> B;
	int lst = Maxl, cnt = 0;
	for (int i = 0; i < A.length(); i++) {
		if (lst == A[i] - 'a') cnt = 0;
		lst = A[i] - 'a'; cnt++;
		for (int j = 0; j < B.length(); j++)
			if (lst != B[j] - 'a')
				dp[i + 1][j + 1][B[j] - 'a'] = (dp[i + 1][j + 1][B[j] - 'a'] + cnt) % mod;
	}
	lst = Maxl, cnt = 0;
	for (int j = 0; j < B.length(); j++) {
		if (lst == B[j] - 'a') cnt = 0;
		lst = B[j] - 'a'; cnt++;
		for (int i = 0; i < A.length(); i++)
			if (lst != A[i] - 'a')
				dp[i + 1][j + 1][A[i] - 'a'] = (dp[i + 1][j + 1][A[i] - 'a'] + cnt) % mod;
	}
	for (int i = 0; i <= A.length(); i++)
		for (int j = 0; j <= B.length(); j++)
			for (int l = 0; l < Maxl; l++) if (dp[i][j][l]) {
				res = (res + dp[i][j][l]) % mod;
				if (i < A.length() && A[i] - 'a' != l)
					dp[i + 1][j][A[i] - 'a'] = (dp[i + 1][j][A[i] - 'a'] + dp[i][j][l]) % mod;
				if (j < B.length() && B[j] - 'a' != l)
					dp[i][j + 1][B[j] - 'a'] = (dp[i][j + 1][B[j] - 'a'] + dp[i][j][l]) % mod;
			}
	printf("%d\n", res);
    return 0;
}

```