#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
 
const int MOD = 998244353;
int N, M;
string x, y;
vector<vector<vector<int>>> dp;
 
inline int add(int a, int b) {
    a += b;
    if (a >= MOD)
        a -= MOD;
    if (a < 0)
        a += MOD;
    return a;
}
 
int main() {
    cin >> x >> y;
    N = x.size(), M = y.size();
    dp.resize(N + 1, vector<vector<int>>(M + 1, vector<int>(2)));
    int ans = 0;
    for (int i = 0; i <= N; i++) {
        for (int j = 0; j <= M; j++) {
            // start substring
            if (i < N)
                dp[i + 1][j][0] = add(dp[i + 1][j][0], 1);
            if (j < M)
                dp[i][j + 1][1] = add(dp[i][j + 1][1], 1);
            // extend
            if (0 < i && i < N && x[i - 1] != x[i])
                dp[i + 1][j][0] = add(dp[i + 1][j][0], dp[i][j][0]);
            if (0 < j && i < N && y[j - 1] != x[i])
                dp[i + 1][j][0] = add(dp[i + 1][j][0], dp[i][j][1]);
            if (0 < i && j < M && x[i - 1] != y[j])
                dp[i][j + 1][1] = add(dp[i][j + 1][1], dp[i][j][0]);
            if (0 < j && j < M && y[j - 1] != y[j])
                dp[i][j + 1][1] = add(dp[i][j + 1][1], dp[i][j][1]);
            // record sum
            ans = add(ans, dp[i][j][0]);
            ans = add(ans, dp[i][j][1]);
        }
    }
    // remove illegal solutions (one substring is empty)
    // ans -= sum(chaotic substrings of x * (M + 1)) + sum(chaotic substrings of y * (N + 1))
    for (int i = 0; i < N;) {
        int j = i + 1, len = 1;
        while (j < N && x[j - 1] != x[j])
            j++, len++;
        ans = ((ans - ll(len) * (len + 1) / 2 % MOD * (M + 1) % MOD) % MOD + MOD) % MOD;
        i = j;
    }
    for (int i = 0; i < M;) {
        int j = i + 1, len = 1;
        while (j < M && y[j - 1] != y[j])
            j++, len++;
        ans = ((ans - ll(len) * (len + 1) / 2 % MOD * (N + 1) % MOD) % MOD + MOD) % MOD;
        i = j;
    }
    cout << ans << endl;
}

### Prompt

Create a solution in cpp for 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 <algorithm>
#include <vector>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
 
const int MOD = 998244353;
int N, M;
string x, y;
vector<vector<vector<int>>> dp;
 
inline int add(int a, int b) {
    a += b;
    if (a >= MOD)
        a -= MOD;
    if (a < 0)
        a += MOD;
    return a;
}
 
int main() {
    cin >> x >> y;
    N = x.size(), M = y.size();
    dp.resize(N + 1, vector<vector<int>>(M + 1, vector<int>(2)));
    int ans = 0;
    for (int i = 0; i <= N; i++) {
        for (int j = 0; j <= M; j++) {
            // start substring
            if (i < N)
                dp[i + 1][j][0] = add(dp[i + 1][j][0], 1);
            if (j < M)
                dp[i][j + 1][1] = add(dp[i][j + 1][1], 1);
            // extend
            if (0 < i && i < N && x[i - 1] != x[i])
                dp[i + 1][j][0] = add(dp[i + 1][j][0], dp[i][j][0]);
            if (0 < j && i < N && y[j - 1] != x[i])
                dp[i + 1][j][0] = add(dp[i + 1][j][0], dp[i][j][1]);
            if (0 < i && j < M && x[i - 1] != y[j])
                dp[i][j + 1][1] = add(dp[i][j + 1][1], dp[i][j][0]);
            if (0 < j && j < M && y[j - 1] != y[j])
                dp[i][j + 1][1] = add(dp[i][j + 1][1], dp[i][j][1]);
            // record sum
            ans = add(ans, dp[i][j][0]);
            ans = add(ans, dp[i][j][1]);
        }
    }
    // remove illegal solutions (one substring is empty)
    // ans -= sum(chaotic substrings of x * (M + 1)) + sum(chaotic substrings of y * (N + 1))
    for (int i = 0; i < N;) {
        int j = i + 1, len = 1;
        while (j < N && x[j - 1] != x[j])
            j++, len++;
        ans = ((ans - ll(len) * (len + 1) / 2 % MOD * (M + 1) % MOD) % MOD + MOD) % MOD;
        i = j;
    }
    for (int i = 0; i < M;) {
        int j = i + 1, len = 1;
        while (j < M && y[j - 1] != y[j])
            j++, len++;
        ans = ((ans - ll(len) * (len + 1) / 2 % MOD * (N + 1) % MOD) % MOD + MOD) % MOD;
        i = j;
    }
    cout << ans << endl;
}

```