#include<bits/stdc++.h>
using namespace std;
const int MOD = 998244353;

int sum(int a, int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int sub(int a, int b) {
    return a - b < 0 ? a - b + MOD : a - b;
}
int mul(int a, int b) {
    return (1LL*a*b)%MOD;
}
int pw(int a, int b) {
	if (!b) return 1;
	int r = pw(a, b/2);
	r = mul(r, r);
	if (b%2) r = mul(r, a);
	return r;
}

const int MAXN = 1003;
string t[2];
int dp[2][MAXN][MAXN];

int solve(int _t, int cur, int other) {
    if (~dp[_t][cur][other]) return dp[_t][cur][other];
    int ans = 1;
    if (cur+1 < t[_t].size() && t[_t][cur] != t[_t][cur+1]) {
        ans = sum(ans, solve(_t, cur+1, other));
    }
    if (other+1 < t[_t^1].size() && t[_t][cur] != t[_t^1][other+1]) {
        ans = sum(ans, solve(_t^1, other+1, cur));
    }
    return dp[_t][cur][other] = ans;
}

int main() {
    cin >> t[0] >> t[1];

    memset(dp, -1, sizeof dp);

    int ans = 0;
    for (int _t : {0, 1}) {
        int run = 0;
        for (int i = 0; i < t[_t].size(); i++) {
            if (i > 0 && t[_t][i-1] == t[_t][i]) run = 0;
            run++;
            for (int j = 0; j < t[_t^1].size(); j++) {
                if (t[_t][i] == t[_t^1][j]) continue;
                ans = sum(ans, mul(run, solve(_t^1, j, i)));
            }
        }
    }

    cout << ans << endl;


    return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;

int sum(int a, int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int sub(int a, int b) {
    return a - b < 0 ? a - b + MOD : a - b;
}
int mul(int a, int b) {
    return (1LL*a*b)%MOD;
}
int pw(int a, int b) {
	if (!b) return 1;
	int r = pw(a, b/2);
	r = mul(r, r);
	if (b%2) r = mul(r, a);
	return r;
}

const int MAXN = 1003;
string t[2];
int dp[2][MAXN][MAXN];

int solve(int _t, int cur, int other) {
    if (~dp[_t][cur][other]) return dp[_t][cur][other];
    int ans = 1;
    if (cur+1 < t[_t].size() && t[_t][cur] != t[_t][cur+1]) {
        ans = sum(ans, solve(_t, cur+1, other));
    }
    if (other+1 < t[_t^1].size() && t[_t][cur] != t[_t^1][other+1]) {
        ans = sum(ans, solve(_t^1, other+1, cur));
    }
    return dp[_t][cur][other] = ans;
}

int main() {
    cin >> t[0] >> t[1];

    memset(dp, -1, sizeof dp);

    int ans = 0;
    for (int _t : {0, 1}) {
        int run = 0;
        for (int i = 0; i < t[_t].size(); i++) {
            if (i > 0 && t[_t][i-1] == t[_t][i]) run = 0;
            run++;
            for (int j = 0; j < t[_t^1].size(); j++) {
                if (t[_t][i] == t[_t^1][j]) continue;
                ans = sum(ans, mul(run, solve(_t^1, j, i)));
            }
        }
    }

    cout << ans << endl;


    return 0;
}

```