/*
ID: espr1t
TASK: 
KEYWORDS: 
*/

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <string>

using namespace std;
FILE* in = stdin; FILE* out = stdout;

const int MAX = 1001;
const int MOD = 998244353;

int n, m;
char a[MAX], b[MAX];
int dyn[MAX][MAX][2];

int recurse(int idxA, int idxB, int last) {
//    fprintf(stderr, "idxA = %d, idxB = %d, last = %d\n", idxA, idxB, last);
    int& ans = dyn[idxA][idxB][last];
    if (ans != -1)
        return ans;

    ans = 1;
    char lastChar = (last ? b[idxB - 1] : a[idxA - 1]);

    // Use a letter in A
    if (idxA < n && a[idxA] != lastChar) {
        ans += recurse(idxA + 1, idxB, 0);
        if (ans >= MOD) ans -= MOD;
    }

    // Use a letter in B
    if (idxB < m && b[idxB] != lastChar) {
        ans += recurse(idxA, idxB + 1, 1);
        if (ans >= MOD) ans -= MOD;
    }
    
    return ans;
}

/*
int recurse(int idxA, int flagA, int idxB, int flagB, int last) {
    int& ans = dyn[idxA][flagA][idxB][flagB][last];
    fprintf(stderr, "idxA = %d, flagA = %d, idxB = %d, flagB = %d, last = %d\n",
        idxA, flagA, idxB, flagB, last);
    if (ans != -1)
        return ans;

    ans = (flagA && flagB);
    char lastChar = (!flagA && !flagB) ? '?' : (last ? b[idxB - 1] : a[idxA - 1]);

    // Skip a letter in A
    if (!flagA && idxA + 1 < n) {
        ans += recurse(idxA + 1, flagA, idxB, flagB, last);
        if (ans >= MOD) ans -= MOD;
    }

    // Skip a letter in B
    if (!flagB && idxB + 1 < m) {
        ans += recurse(idxA, flagA, idxB + 1, flagB, last);
        if (ans >= MOD) ans -= MOD;
    }

    // Use a letter in A
    if (idxA < n && a[idxA] != lastChar) {
        ans += recurse(idxA + 1, 1, idxB, flagB, 0);
        if (ans >= MOD) ans -= MOD;
    }

    // Use a letter in B
    if (idxB < m && b[idxB] != lastChar) {
        ans += recurse(idxA, flagA, idxB + 1, 1, 1);
        if (ans >= MOD) ans -= MOD;
    }
    
    return ans;
}
*/

int solve() {
    memset(dyn, -1, sizeof(dyn));
    
    int ans = 0;
    for (int lastA = 0; lastA < n; lastA++) {
        int cntBefore = 0;
        while (lastA - cntBefore - 1 >= 0 && a[lastA - cntBefore - 1] != a[lastA - cntBefore])
            cntBefore++;
        for (int startB = 0; startB < m; startB++) {
            if (a[lastA] != b[startB]) {
                ans = (ans + (long long)(cntBefore + 1) * recurse(lastA + 1, startB + 1, 1)) % MOD;
            }
        }
    }
    for (int lastB = 0; lastB < m; lastB++) {
        int cntBefore = 0;
        while (lastB - cntBefore - 1 >= 0 && b[lastB - cntBefore - 1] != b[lastB - cntBefore])
            cntBefore++;
        for (int startA = 0; startA < n; startA++) {
            if (a[startA] != b[lastB]) {
                ans = (ans + (long long)(cntBefore + 1) * recurse(startA + 1, lastB + 1, 0)) % MOD;
            }
        }
    }
    return ans;
}

int main(void) {
	// in = fopen("E.in", "rt");
	
	fscanf(in, "%s", a); n = (int)strlen(a);
	fscanf(in, "%s", b); m = (int)strlen(b);
	fprintf(out, "%d\n", solve());
	
	return 0;
}

### Prompt

In CPP, your task is to solve 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
/*
ID: espr1t
TASK: 
KEYWORDS: 
*/

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <string>

using namespace std;
FILE* in = stdin; FILE* out = stdout;

const int MAX = 1001;
const int MOD = 998244353;

int n, m;
char a[MAX], b[MAX];
int dyn[MAX][MAX][2];

int recurse(int idxA, int idxB, int last) {
//    fprintf(stderr, "idxA = %d, idxB = %d, last = %d\n", idxA, idxB, last);
    int& ans = dyn[idxA][idxB][last];
    if (ans != -1)
        return ans;

    ans = 1;
    char lastChar = (last ? b[idxB - 1] : a[idxA - 1]);

    // Use a letter in A
    if (idxA < n && a[idxA] != lastChar) {
        ans += recurse(idxA + 1, idxB, 0);
        if (ans >= MOD) ans -= MOD;
    }

    // Use a letter in B
    if (idxB < m && b[idxB] != lastChar) {
        ans += recurse(idxA, idxB + 1, 1);
        if (ans >= MOD) ans -= MOD;
    }
    
    return ans;
}

/*
int recurse(int idxA, int flagA, int idxB, int flagB, int last) {
    int& ans = dyn[idxA][flagA][idxB][flagB][last];
    fprintf(stderr, "idxA = %d, flagA = %d, idxB = %d, flagB = %d, last = %d\n",
        idxA, flagA, idxB, flagB, last);
    if (ans != -1)
        return ans;

    ans = (flagA && flagB);
    char lastChar = (!flagA && !flagB) ? '?' : (last ? b[idxB - 1] : a[idxA - 1]);

    // Skip a letter in A
    if (!flagA && idxA + 1 < n) {
        ans += recurse(idxA + 1, flagA, idxB, flagB, last);
        if (ans >= MOD) ans -= MOD;
    }

    // Skip a letter in B
    if (!flagB && idxB + 1 < m) {
        ans += recurse(idxA, flagA, idxB + 1, flagB, last);
        if (ans >= MOD) ans -= MOD;
    }

    // Use a letter in A
    if (idxA < n && a[idxA] != lastChar) {
        ans += recurse(idxA + 1, 1, idxB, flagB, 0);
        if (ans >= MOD) ans -= MOD;
    }

    // Use a letter in B
    if (idxB < m && b[idxB] != lastChar) {
        ans += recurse(idxA, flagA, idxB + 1, 1, 1);
        if (ans >= MOD) ans -= MOD;
    }
    
    return ans;
}
*/

int solve() {
    memset(dyn, -1, sizeof(dyn));
    
    int ans = 0;
    for (int lastA = 0; lastA < n; lastA++) {
        int cntBefore = 0;
        while (lastA - cntBefore - 1 >= 0 && a[lastA - cntBefore - 1] != a[lastA - cntBefore])
            cntBefore++;
        for (int startB = 0; startB < m; startB++) {
            if (a[lastA] != b[startB]) {
                ans = (ans + (long long)(cntBefore + 1) * recurse(lastA + 1, startB + 1, 1)) % MOD;
            }
        }
    }
    for (int lastB = 0; lastB < m; lastB++) {
        int cntBefore = 0;
        while (lastB - cntBefore - 1 >= 0 && b[lastB - cntBefore - 1] != b[lastB - cntBefore])
            cntBefore++;
        for (int startA = 0; startA < n; startA++) {
            if (a[startA] != b[lastB]) {
                ans = (ans + (long long)(cntBefore + 1) * recurse(startA + 1, lastB + 1, 0)) % MOD;
            }
        }
    }
    return ans;
}

int main(void) {
	// in = fopen("E.in", "rt");
	
	fscanf(in, "%s", a); n = (int)strlen(a);
	fscanf(in, "%s", b); m = (int)strlen(b);
	fprintf(out, "%d\n", solve());
	
	return 0;
}

```