#include <iostream>
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e3+5;
long long dp[MAXN][MAXN][5];
int a[MAXN];
int b[MAXN];
long long ans;
long long MOD = 998244353;
int main() {
    string a,b;
    cin>>a>>b;
    int n = a.length();
    int m = b.length();
    for(int i=0;i<=n;i++){
        for(int j=0;j<=m;j++){
            if(i){
                dp[i][j][0] = 1;
                if(i-2<0 || a[i-1]!=a[i-2]){
                    dp[i][j][0] += dp[i-1][j][0];
                    dp[i][j][0]%=MOD;
                }
                if(j-1<0 || a[i-1]!=b[j-1]){
                    dp[i][j][0] += dp[i-1][j][1];
                    dp[i][j][0]%=MOD;
                }
            }
            if(j){
                dp[i][j][1] =1;
                if(j-2<0 || b[j-1]!=b[j-2]){
                    dp[i][j][1] += dp[i][j-1][1];
                    dp[i][j][1]%=MOD;
                }
                if(i-1<0 || a[i-1]!=b[j-1]){
                    dp[i][j][1] += dp[i][j-1][0];
                    dp[i][j][1]%=MOD;
                }
            }
            ans+=dp[i][j][0]+dp[i][j][1];
            ans%=MOD;
        }
    }
   
    for(int i=1;i<=n;i++){
        for(int j=i;j<=n;j++){
            if(j>i && a[j-1]==a[j-2]){
                break;
            }
            ans=(ans-(m+1)+MOD)%MOD;
        }
    }
    
    for(int i=1;i<=m;i++){
        for(int j=i;j<=m;j++){
            if(j>i && b[j-1]==b[j-2]){
                break;
            }
            ans=(ans-(n+1)+MOD)%MOD;
        }
    }
    cout<<ans<<endl;
}
### Prompt

Please create a solution in Cpp 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 <iostream>
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e3+5;
long long dp[MAXN][MAXN][5];
int a[MAXN];
int b[MAXN];
long long ans;
long long MOD = 998244353;
int main() {
    string a,b;
    cin>>a>>b;
    int n = a.length();
    int m = b.length();
    for(int i=0;i<=n;i++){
        for(int j=0;j<=m;j++){
            if(i){
                dp[i][j][0] = 1;
                if(i-2<0 || a[i-1]!=a[i-2]){
                    dp[i][j][0] += dp[i-1][j][0];
                    dp[i][j][0]%=MOD;
                }
                if(j-1<0 || a[i-1]!=b[j-1]){
                    dp[i][j][0] += dp[i-1][j][1];
                    dp[i][j][0]%=MOD;
                }
            }
            if(j){
                dp[i][j][1] =1;
                if(j-2<0 || b[j-1]!=b[j-2]){
                    dp[i][j][1] += dp[i][j-1][1];
                    dp[i][j][1]%=MOD;
                }
                if(i-1<0 || a[i-1]!=b[j-1]){
                    dp[i][j][1] += dp[i][j-1][0];
                    dp[i][j][1]%=MOD;
                }
            }
            ans+=dp[i][j][0]+dp[i][j][1];
            ans%=MOD;
        }
    }
   
    for(int i=1;i<=n;i++){
        for(int j=i;j<=n;j++){
            if(j>i && a[j-1]==a[j-2]){
                break;
            }
            ans=(ans-(m+1)+MOD)%MOD;
        }
    }
    
    for(int i=1;i<=m;i++){
        for(int j=i;j<=m;j++){
            if(j>i && b[j-1]==b[j-2]){
                break;
            }
            ans=(ans-(n+1)+MOD)%MOD;
        }
    }
    cout<<ans<<endl;
}
```