#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int,int> pii;
const int MOD = 998244353;

char s[1010], t[1010];
int n, m;
int f[1010][1010][2];

void upd(int &x, int y) {
    x += y;
    if (x >= MOD) x -= MOD;
}

int main() {
    scanf("%s", s + 1);
    scanf("%s", t + 1);
    n = strlen(s + 1);
    m = strlen(t + 1);
    memset(f, 0, sizeof(f));
    for (int i=1;i<=m;i++) {
        f[0][i][1] = 1;
        if (i > 0 && t[i] != t[i - 1]) upd(f[0][i][1], f[0][i - 1][1]);
    }
    for (int i=1;i<=n;i++) {
        f[i][0][0] = 1;
        if (i > 0 && s[i] != s[i - 1]) upd(f[i][0][0], f[i - 1][0][0]);
    }
    int ans = 0;
    for (int i=1;i<=n;i++) {
        for (int j=1;j<=m;j++) {
            // push i
            f[i][j][0] = 0;
            if (s[i] != t[j]) {
                upd(f[i][j][0], f[0][j][1]);
                if (i > 1) upd(f[i][j][0], f[i - 1][j][1]);
            }
            if (s[i] != s[i - 1] && i > 1) {
                upd(f[i][j][0], f[i - 1][j][0]);
            }
            // push j
            if (s[i] != t[j]) {
                upd(f[i][j][1], f[i][0][0]);
                if (j > 1) upd(f[i][j][1], f[i][j - 1][0]);
            }
            if (t[j] != t[j - 1] && j > 1) {
                upd(f[i][j][1], f[i][j - 1][1]);
            }
            upd(ans, f[i][j][0]);
            upd(ans, f[i][j][1]);
        }
    }
    printf("%d\n",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;
typedef pair<int,int> pii;
const int MOD = 998244353;

char s[1010], t[1010];
int n, m;
int f[1010][1010][2];

void upd(int &x, int y) {
    x += y;
    if (x >= MOD) x -= MOD;
}

int main() {
    scanf("%s", s + 1);
    scanf("%s", t + 1);
    n = strlen(s + 1);
    m = strlen(t + 1);
    memset(f, 0, sizeof(f));
    for (int i=1;i<=m;i++) {
        f[0][i][1] = 1;
        if (i > 0 && t[i] != t[i - 1]) upd(f[0][i][1], f[0][i - 1][1]);
    }
    for (int i=1;i<=n;i++) {
        f[i][0][0] = 1;
        if (i > 0 && s[i] != s[i - 1]) upd(f[i][0][0], f[i - 1][0][0]);
    }
    int ans = 0;
    for (int i=1;i<=n;i++) {
        for (int j=1;j<=m;j++) {
            // push i
            f[i][j][0] = 0;
            if (s[i] != t[j]) {
                upd(f[i][j][0], f[0][j][1]);
                if (i > 1) upd(f[i][j][0], f[i - 1][j][1]);
            }
            if (s[i] != s[i - 1] && i > 1) {
                upd(f[i][j][0], f[i - 1][j][0]);
            }
            // push j
            if (s[i] != t[j]) {
                upd(f[i][j][1], f[i][0][0]);
                if (j > 1) upd(f[i][j][1], f[i][j - 1][0]);
            }
            if (t[j] != t[j - 1] && j > 1) {
                upd(f[i][j][1], f[i][j - 1][1]);
            }
            upd(ans, f[i][j][0]);
            upd(ans, f[i][j][1]);
        }
    }
    printf("%d\n",ans);
}
```