#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int N = 1010;
const int mod = 998244353;

int dp[N][N][2];
string x, y;

void add(int& a, int b){
  a += b;
  if(a >= mod)
    a -= mod;
}

int main(){
  ios_base::sync_with_stdio(0);
  cin.tie(0); cout.tie(0);
  cin >> x >> y;
  int n = x.size(), m = y.size();
  for(int i = 0; i < n; i ++){
    for(int j = i; j < n; j ++){
      if(j > i && x[j] == x[j - 1])
        break;
      for(int k = 0; k < m; k ++){
        if(x[j] != y[k])
          dp[j][k][1] = max(dp[j][k][1], (j - i + 1));
      }
    }
  }
  for(int i = 0; i < m; i ++){
    for(int j = i; j < m; j ++){
      if(j > i && y[j] == y[j - 1])
        break;
      for(int k = 0; k < n; k ++){
        if(y[j] != x[k])
          dp[k][j][0] = max(dp[k][j][0], (j - i + 1));
      }
    }
  }
//  for(int i = 0; i < n; i ++)
//    for(int j = 0; j < m; j ++)
//      if(x[i] != y[j])
//        dp[i][j][0] = dp[i][j][1] = 1;
  for(int i = 0; i < n; i ++){
    for(int j = 0; j < m; j ++){
      if(i + 1 < n){
        if(x[i + 1] != x[i])
          add(dp[i + 1][j][0], dp[i][j][0]);
        if(x[i + 1] != y[j])
          add(dp[i + 1][j][0], dp[i][j][1]);
      }
      if(j + 1 < m){
        if(y[j + 1] != y[j])
          add(dp[i][j + 1][1], dp[i][j][1]);
        if(y[j + 1] != x[i])
          add(dp[i][j + 1][1], dp[i][j][0]);
      }
    }
  }
  int ans = 0;
  for(int i = 0; i < n; i ++)
    for(int j = 0; j < m; j ++)
      for(int k = 0; k < 2; k ++)
        add(ans, dp[i][j][k]);
  cout << ans;
}

### Prompt

Generate 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;
typedef long long ll;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int N = 1010;
const int mod = 998244353;

int dp[N][N][2];
string x, y;

void add(int& a, int b){
  a += b;
  if(a >= mod)
    a -= mod;
}

int main(){
  ios_base::sync_with_stdio(0);
  cin.tie(0); cout.tie(0);
  cin >> x >> y;
  int n = x.size(), m = y.size();
  for(int i = 0; i < n; i ++){
    for(int j = i; j < n; j ++){
      if(j > i && x[j] == x[j - 1])
        break;
      for(int k = 0; k < m; k ++){
        if(x[j] != y[k])
          dp[j][k][1] = max(dp[j][k][1], (j - i + 1));
      }
    }
  }
  for(int i = 0; i < m; i ++){
    for(int j = i; j < m; j ++){
      if(j > i && y[j] == y[j - 1])
        break;
      for(int k = 0; k < n; k ++){
        if(y[j] != x[k])
          dp[k][j][0] = max(dp[k][j][0], (j - i + 1));
      }
    }
  }
//  for(int i = 0; i < n; i ++)
//    for(int j = 0; j < m; j ++)
//      if(x[i] != y[j])
//        dp[i][j][0] = dp[i][j][1] = 1;
  for(int i = 0; i < n; i ++){
    for(int j = 0; j < m; j ++){
      if(i + 1 < n){
        if(x[i + 1] != x[i])
          add(dp[i + 1][j][0], dp[i][j][0]);
        if(x[i + 1] != y[j])
          add(dp[i + 1][j][0], dp[i][j][1]);
      }
      if(j + 1 < m){
        if(y[j + 1] != y[j])
          add(dp[i][j + 1][1], dp[i][j][1]);
        if(y[j + 1] != x[i])
          add(dp[i][j + 1][1], dp[i][j][0]);
      }
    }
  }
  int ans = 0;
  for(int i = 0; i < n; i ++)
    for(int j = 0; j < m; j ++)
      for(int k = 0; k < 2; k ++)
        add(ans, dp[i][j][k]);
  cout << ans;
}

```