#include <bits/stdc++.h>
using namespace std;
 
#define S second
#define LL long long
#define F first

const int N = 1e3+10;

LL LINF = 1e18;
LL INF = 1e9;
LL MOD = 998244353;

string s[2];
int dp[N][N][30];

int add(int a,int b){
    return (a+b)%MOD;
}

int main(){

    //freopen("out.txt","w",stdout);  

    cin>>s[0]>>s[1];
    s[0] = '*'+s[0];
    s[1] = '*'+s[1];

    dp[0][0][26] = 1;

    for(int i=0;i<s[0].size();i++)
        for(int j=0;j<s[1].size();j++){
            
            if(i)
            for(int k=0;k<=26;k++)
                if(k!=s[0][i]-'a'){
                    dp[i][j][s[0][i]-'a'] = add(dp[i][j][s[0][i]-'a'],dp[0][j][k]);
                    if(i>1)dp[i][j][s[0][i]-'a'] = add(dp[i][j][s[0][i]-'a'],dp[i-1][j][k]);
                }

            if(j)
            for(int k=0;k<=26;k++)
                if(k!=s[1][j]-'a'){
                    dp[i][j][s[1][j]-'a'] = add(dp[i][j][s[1][j]-'a'],dp[i][0][k]);
                    if(j>1)dp[i][j][s[1][j]-'a'] = add(dp[i][j][s[1][j]-'a'],dp[i][j-1][k]);
                }
        }

    LL ans = 0;
    for(int i=1;i<s[0].size();i++)
        for(int j=1;j<s[1].size();j++)
            for(int k=0;k<26;k++)
                ans = add(ans,dp[i][j][k]);
    
    printf("%lld\n",ans);
}       
### 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;
 
#define S second
#define LL long long
#define F first

const int N = 1e3+10;

LL LINF = 1e18;
LL INF = 1e9;
LL MOD = 998244353;

string s[2];
int dp[N][N][30];

int add(int a,int b){
    return (a+b)%MOD;
}

int main(){

    //freopen("out.txt","w",stdout);  

    cin>>s[0]>>s[1];
    s[0] = '*'+s[0];
    s[1] = '*'+s[1];

    dp[0][0][26] = 1;

    for(int i=0;i<s[0].size();i++)
        for(int j=0;j<s[1].size();j++){
            
            if(i)
            for(int k=0;k<=26;k++)
                if(k!=s[0][i]-'a'){
                    dp[i][j][s[0][i]-'a'] = add(dp[i][j][s[0][i]-'a'],dp[0][j][k]);
                    if(i>1)dp[i][j][s[0][i]-'a'] = add(dp[i][j][s[0][i]-'a'],dp[i-1][j][k]);
                }

            if(j)
            for(int k=0;k<=26;k++)
                if(k!=s[1][j]-'a'){
                    dp[i][j][s[1][j]-'a'] = add(dp[i][j][s[1][j]-'a'],dp[i][0][k]);
                    if(j>1)dp[i][j][s[1][j]-'a'] = add(dp[i][j][s[1][j]-'a'],dp[i][j-1][k]);
                }
        }

    LL ans = 0;
    for(int i=1;i<s[0].size();i++)
        for(int j=1;j<s[1].size();j++)
            for(int k=0;k<26;k++)
                ans = add(ans,dp[i][j][k]);
    
    printf("%lld\n",ans);
}       
```