#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const int MOD = 998244353;
string a, b;
int dp[1010][1010][2];

int add(int a, int b){
    int ret = a+b;
    if (ret>=MOD) return ret-MOD;
    else return ret;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cin >> a >> b;
    int n = a.size(), m = b.size();
    for (int i=1;i<=n;i++){
        dp[i][0][1] = 0;
        dp[i][0][0] = 1;
        if (i>1 && a[i-1]!=a[i-2]) dp[i][0][0] = add(dp[i][0][0], dp[i-1][0][0]);
        //printf("%d: %d\n", i, dp[i][0][0]);
    }
    int ans=0;
    for (int j=1;j<=m;j++){
        dp[0][j][0] = 0;
        dp[0][j][1] = 1;
        if (j>1 && b[j-1]!=b[j-2]) dp[0][j][1] = add(dp[0][j][1], dp[0][j-1][1]);
        for (int i=1;i<=n;i++){
            if (a[i-1]!=b[j-1]) dp[i][j][0] = add(dp[0][j][1], dp[i-1][j][1]), dp[i][j][1] = add(dp[i][0][0], dp[i][j-1][0]);
            if (i==1 && a[i-1]!=b[j-1]) dp[i][j][0] = (dp[i][j][0]+MOD-dp[0][j][1])%MOD;
            if (j==1 && a[i-1]!=b[j-1]) dp[i][j][1] = (dp[i][j][1]+MOD-dp[i][0][0])%MOD;
            if (i>1 && a[i-1]!=a[i-2]) dp[i][j][0] = add(dp[i][j][0], dp[i-1][j][0]);
            if (j>1 && b[j-1]!=b[j-2]) dp[i][j][1] = add(dp[i][j][1], dp[i][j-1][1]);
            //printf("%d %d: %d %d\n", i, j, dp[i][j][0], dp[i][j][1]);
            ans = add(ans, add(dp[i][j][0], dp[i][j][1]));
        }
    }
    printf("%d\n", ans);
    return 0;
}

### Prompt

Create a solution in Cpp for 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;
typedef long long ll;
const int MOD = 998244353;
string a, b;
int dp[1010][1010][2];

int add(int a, int b){
    int ret = a+b;
    if (ret>=MOD) return ret-MOD;
    else return ret;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cin >> a >> b;
    int n = a.size(), m = b.size();
    for (int i=1;i<=n;i++){
        dp[i][0][1] = 0;
        dp[i][0][0] = 1;
        if (i>1 && a[i-1]!=a[i-2]) dp[i][0][0] = add(dp[i][0][0], dp[i-1][0][0]);
        //printf("%d: %d\n", i, dp[i][0][0]);
    }
    int ans=0;
    for (int j=1;j<=m;j++){
        dp[0][j][0] = 0;
        dp[0][j][1] = 1;
        if (j>1 && b[j-1]!=b[j-2]) dp[0][j][1] = add(dp[0][j][1], dp[0][j-1][1]);
        for (int i=1;i<=n;i++){
            if (a[i-1]!=b[j-1]) dp[i][j][0] = add(dp[0][j][1], dp[i-1][j][1]), dp[i][j][1] = add(dp[i][0][0], dp[i][j-1][0]);
            if (i==1 && a[i-1]!=b[j-1]) dp[i][j][0] = (dp[i][j][0]+MOD-dp[0][j][1])%MOD;
            if (j==1 && a[i-1]!=b[j-1]) dp[i][j][1] = (dp[i][j][1]+MOD-dp[i][0][0])%MOD;
            if (i>1 && a[i-1]!=a[i-2]) dp[i][j][0] = add(dp[i][j][0], dp[i-1][j][0]);
            if (j>1 && b[j-1]!=b[j-2]) dp[i][j][1] = add(dp[i][j][1], dp[i][j-1][1]);
            //printf("%d %d: %d %d\n", i, j, dp[i][j][0], dp[i][j][1]);
            ans = add(ans, add(dp[i][j][0], dp[i][j][1]));
        }
    }
    printf("%d\n", ans);
    return 0;
}

```