#include <bits/stdc++.h>

using namespace std;
const int c=1002;
long long n, m, dp[c][c][2], sum=0, mod=998244353;
string a, b;
int main()
{
    ios_base::sync_with_stdio(false);
    cin >> a >> b;
    n=a.size(), m=b.size();
    a+='_', b+='_';
    for (int i=n-1; i>=0; i--) {
        if (a[i]!=a[i+1]) {
            dp[i][m][0]=dp[i+1][m][0];
        }
        dp[i][m][0]++;
    }
    for (int j=m-1; j>=0; j--) {
        if (b[j]!=b[j+1]) {
            dp[n][j][1]+=dp[n][j+1][1];
        }
        dp[n][j][1]++;
    }
    for (int i=n-1; i>=0; i--) {
        for (int j=m-1; j>=0; j--) {
            long long x=0, y=0;
            if (a[i]!=b[j]) {
                x+=dp[i+1][j][1];
                if (i!=n-1) x+=dp[n][j][1];
                y+=dp[i][j+1][0];
                if (j!=m-1) y+=dp[i][m][0];
            }
            if (a[i]!=a[i+1] && i!=n-1) {
                x+=dp[i+1][j][0];
            }
            if (b[j]!=b[j+1] && j!=m-1) {
                y+=dp[i][j+1][1];
            }
            x%=mod, y%=mod;
            dp[i][j][0]=x, dp[i][j][1]=y;
        }
    }
    for (int i=0; i<n; i++) {
        for (int j=0; j<m; j++) {
            //cout << i << " " << j << " " << dp[i][j][0] << " " << dp[i][j][1] << "\n";
            sum+=dp[i][j][0]+dp[i][j][1];
            sum%=mod;
        }
    }
    cout << sum << "\n";
    return 0;
}
/*
aa
b
*/

### 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 c=1002;
long long n, m, dp[c][c][2], sum=0, mod=998244353;
string a, b;
int main()
{
    ios_base::sync_with_stdio(false);
    cin >> a >> b;
    n=a.size(), m=b.size();
    a+='_', b+='_';
    for (int i=n-1; i>=0; i--) {
        if (a[i]!=a[i+1]) {
            dp[i][m][0]=dp[i+1][m][0];
        }
        dp[i][m][0]++;
    }
    for (int j=m-1; j>=0; j--) {
        if (b[j]!=b[j+1]) {
            dp[n][j][1]+=dp[n][j+1][1];
        }
        dp[n][j][1]++;
    }
    for (int i=n-1; i>=0; i--) {
        for (int j=m-1; j>=0; j--) {
            long long x=0, y=0;
            if (a[i]!=b[j]) {
                x+=dp[i+1][j][1];
                if (i!=n-1) x+=dp[n][j][1];
                y+=dp[i][j+1][0];
                if (j!=m-1) y+=dp[i][m][0];
            }
            if (a[i]!=a[i+1] && i!=n-1) {
                x+=dp[i+1][j][0];
            }
            if (b[j]!=b[j+1] && j!=m-1) {
                y+=dp[i][j+1][1];
            }
            x%=mod, y%=mod;
            dp[i][j][0]=x, dp[i][j][1]=y;
        }
    }
    for (int i=0; i<n; i++) {
        for (int j=0; j<m; j++) {
            //cout << i << " " << j << " " << dp[i][j][0] << " " << dp[i][j][1] << "\n";
            sum+=dp[i][j][0]+dp[i][j][1];
            sum%=mod;
        }
    }
    cout << sum << "\n";
    return 0;
}
/*
aa
b
*/

```