
#include <bits/stdc++.h>

#define ll long long
#define sz(x) ((int) (x).size())
#define all(x) (x).begin(), (x).end()
#define vi vector<int>
#define pii pair<int, int>
#define rep(i, a, b) for(int i = (a); i < (b); i++)
using namespace std;
template<typename T>
using minpq = priority_queue<T, vector<T>, greater<T>>;

const int M = 998244353;
const int N = 1005;

ll dp[N][N][3];

ll cnt(const string &s) {
    int j = 1;
    int n = s.length();
    ll ans = 0;
    rep(i, 1, n) {
        j = max(j, i + 1);
        while(j < n && s[j] != s[j - 1]) j++;
        ans = (ans + j - i) % M;
    }
    return ans;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    string x, y;
    cin >> x >> y;
    x = "$" + x;
    y = "$" + y;
    int n = x.length(), m = y.length();
    ll ans = 0;
    for(int i = n; i >= 1; i--) for(int j = m; j >= 1; j--) {
        dp[i][j][0] = dp[i][j][1] = dp[i][j][2] = 1;
        if(i < n) {
            dp[i][j][0] = (dp[i][j][0] + dp[i + 1][j][1]) % M;
            if(x[i] != x[i - 1]) {
                dp[i][j][1] = (dp[i][j][1] + dp[i + 1][j][1]) % M;
            }
            if(x[i] != y[j - 1]) {
                dp[i][j][2] = (dp[i][j][2] + dp[i + 1][j][1]) % M;
            }
        }
        if(j < m) {
            dp[i][j][0] = (dp[i][j][0] + dp[i][j + 1][2]) % M;
            if(y[j] != x[i - 1]) {
                dp[i][j][1] = (dp[i][j][1] + dp[i][j + 1][2]) % M;
            }
            if(y[j] != y[j - 1]) {
                dp[i][j][2] = (dp[i][j][2] + dp[i][j + 1][2]) % M;
            }
        }
        // cout << i << ' ' << j << ":\t" << dp[i][j][0] << ' ' << dp[i][j][1] << ' ' << dp[i][j][2] << '\n';
        ans = (ans + dp[i][j][0] - 1) % M;
    }
    ll sub = (cnt(x) * m) % M;
    sub = (sub + cnt(y) * n) % M;
    ans -= sub;

    ans = (ans + M) % M;
    cout << ans << '\n';
}
### 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>

#define ll long long
#define sz(x) ((int) (x).size())
#define all(x) (x).begin(), (x).end()
#define vi vector<int>
#define pii pair<int, int>
#define rep(i, a, b) for(int i = (a); i < (b); i++)
using namespace std;
template<typename T>
using minpq = priority_queue<T, vector<T>, greater<T>>;

const int M = 998244353;
const int N = 1005;

ll dp[N][N][3];

ll cnt(const string &s) {
    int j = 1;
    int n = s.length();
    ll ans = 0;
    rep(i, 1, n) {
        j = max(j, i + 1);
        while(j < n && s[j] != s[j - 1]) j++;
        ans = (ans + j - i) % M;
    }
    return ans;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    string x, y;
    cin >> x >> y;
    x = "$" + x;
    y = "$" + y;
    int n = x.length(), m = y.length();
    ll ans = 0;
    for(int i = n; i >= 1; i--) for(int j = m; j >= 1; j--) {
        dp[i][j][0] = dp[i][j][1] = dp[i][j][2] = 1;
        if(i < n) {
            dp[i][j][0] = (dp[i][j][0] + dp[i + 1][j][1]) % M;
            if(x[i] != x[i - 1]) {
                dp[i][j][1] = (dp[i][j][1] + dp[i + 1][j][1]) % M;
            }
            if(x[i] != y[j - 1]) {
                dp[i][j][2] = (dp[i][j][2] + dp[i + 1][j][1]) % M;
            }
        }
        if(j < m) {
            dp[i][j][0] = (dp[i][j][0] + dp[i][j + 1][2]) % M;
            if(y[j] != x[i - 1]) {
                dp[i][j][1] = (dp[i][j][1] + dp[i][j + 1][2]) % M;
            }
            if(y[j] != y[j - 1]) {
                dp[i][j][2] = (dp[i][j][2] + dp[i][j + 1][2]) % M;
            }
        }
        // cout << i << ' ' << j << ":\t" << dp[i][j][0] << ' ' << dp[i][j][1] << ' ' << dp[i][j][2] << '\n';
        ans = (ans + dp[i][j][0] - 1) % M;
    }
    ll sub = (cnt(x) * m) % M;
    sub = (sub + cnt(y) * n) % M;
    ans -= sub;

    ans = (ans + M) % M;
    cout << ans << '\n';
}
```