#include <bits/stdc++.h>
#define debug(x) cout << #x << " = " << x << endl
#define REP(i, n) for (Long i = 0; i < (Long)n; i++)
using namespace std;

typedef long long Long;

string x, y;

const int MX = 1005;
bool used[MX][MX][2][2][2];
Long dp[MX][MX][2][2][2];

const Long MOD = 998244353;
Long add(Long a, Long b) {
	if (a + b < MOD) return a + b;
	return a + b - MOD;
}

Long count(Long i, Long j, Long prev, bool left = false, bool right = false, char prevChar = '$') {
	if (i == x.size() && j == y.size()) {
		return left && right;
	}
	Long &ans = dp[i][j][prev][left][right];
	if (used[i][j][prev][left][right]) {
		return ans;
	}
	used[i][j][prev][left][right] = true;
	ans = 0;
	if (left && right) ans++;
	if (i < x.size() && x[i] != prevChar) {
		ans = add(ans, count(i + 1, j, 0, true, right , x[i]));
	}
	if (j < y.size() && y[j] != prevChar) {
		ans = add(ans, count(i, j + 1, 1, left, true, y[j]));
	}
	return ans;
}

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);

	cin >> x >> y;
	
	Long ans = 0;
	for (int l1 = 0; l1 < x.size(); l1++) {
		for (int l2 = 0; l2 < y.size(); l2++) {
			ans = add(ans, count(l1, l2, 0));
		}
	}
	cout << ans << "\n";
	
	return 0;
}

### Prompt

In cpp, your task is to solve 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>
#define debug(x) cout << #x << " = " << x << endl
#define REP(i, n) for (Long i = 0; i < (Long)n; i++)
using namespace std;

typedef long long Long;

string x, y;

const int MX = 1005;
bool used[MX][MX][2][2][2];
Long dp[MX][MX][2][2][2];

const Long MOD = 998244353;
Long add(Long a, Long b) {
	if (a + b < MOD) return a + b;
	return a + b - MOD;
}

Long count(Long i, Long j, Long prev, bool left = false, bool right = false, char prevChar = '$') {
	if (i == x.size() && j == y.size()) {
		return left && right;
	}
	Long &ans = dp[i][j][prev][left][right];
	if (used[i][j][prev][left][right]) {
		return ans;
	}
	used[i][j][prev][left][right] = true;
	ans = 0;
	if (left && right) ans++;
	if (i < x.size() && x[i] != prevChar) {
		ans = add(ans, count(i + 1, j, 0, true, right , x[i]));
	}
	if (j < y.size() && y[j] != prevChar) {
		ans = add(ans, count(i, j + 1, 1, left, true, y[j]));
	}
	return ans;
}

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);

	cin >> x >> y;
	
	Long ans = 0;
	for (int l1 = 0; l1 < x.size(); l1++) {
		for (int l2 = 0; l2 < y.size(); l2++) {
			ans = add(ans, count(l1, l2, 0));
		}
	}
	cout << ans << "\n";
	
	return 0;
}

```