#include <cstdio>
#define MOD 998244353

char A[1010],B[1010];
long long ans;

long long f[1010][1010][2],g[1010],h[1010];

int main()
{
    scanf("%s%s",A+1,B+1);
    for(int i=1;A[i];i++) g[i] = A[i] == A[i-1] ? 1 : g[i-1] + 1;
    for(int i=1;B[i];i++) h[i] = B[i] == B[i-1] ? 1 : h[i-1] + 1;

    #define ADD(a, b) a = (a + (b)) % MOD
    for(int i=1;A[i];i++) for(int j=1;B[j];j++)
    {
        if(A[i] != A[i-1]) ADD(f[i][j][0], f[i-1][j][0]);
        if(B[j] != B[j-1]) ADD(f[i][j][1], f[i][j-1][1]);
        if(A[i] != B[j]) ADD(f[i][j][0], f[i-1][j][1] + h[j]);
        if(A[i] != B[j]) ADD(f[i][j][1], f[i][j-1][0] + g[i]);
        ADD(ans, f[i][j][0] + f[i][j][1]);
    }
    printf("%lld\n",ans);
    return 0;
}

### Prompt

Please formulate 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 <cstdio>
#define MOD 998244353

char A[1010],B[1010];
long long ans;

long long f[1010][1010][2],g[1010],h[1010];

int main()
{
    scanf("%s%s",A+1,B+1);
    for(int i=1;A[i];i++) g[i] = A[i] == A[i-1] ? 1 : g[i-1] + 1;
    for(int i=1;B[i];i++) h[i] = B[i] == B[i-1] ? 1 : h[i-1] + 1;

    #define ADD(a, b) a = (a + (b)) % MOD
    for(int i=1;A[i];i++) for(int j=1;B[j];j++)
    {
        if(A[i] != A[i-1]) ADD(f[i][j][0], f[i-1][j][0]);
        if(B[j] != B[j-1]) ADD(f[i][j][1], f[i][j-1][1]);
        if(A[i] != B[j]) ADD(f[i][j][0], f[i-1][j][1] + h[j]);
        if(A[i] != B[j]) ADD(f[i][j][1], f[i][j-1][0] + g[i]);
        ADD(ans, f[i][j][0] + f[i][j][1]);
    }
    printf("%lld\n",ans);
    return 0;
}

```