#include <bits/stdc++.h>

using namespace std;

#define mem(a,b) memset(a,b,sizeof(a))
#define FOR(i,j,k) for(int i=j;i<=k;i++)
#define read        freopen("in.txt", "r", stdin)
#define write       freopen("out.txt", "w", stdout)
#define pf          printf
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define PB push_back
#define F first
#define S second
#define MP make_pair

typedef long long ll;
typedef pair<int,int> pii;

const int M = 998244353 ;
const int MAX = (1e3) + 10;

mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());

int n, m, dp[MAX][MAX][4][2];
string A, B;

int add(int x,int y) {
    int r = x + y;
    return (r >= M) ? (r - M) : r;
}

int rec(int x,int y,int s,int t) {
    // cout << x << " " << y << ' ' << s << " " << t << endl;
    int &temp = dp[x][y][s][t];
    if(temp != -1) return temp;

    temp = s == 3 ? 1 : 0;
    char c = t == 0 ? A[x - 1] : B[y - 1];

    if(x < n && A[x] != c) {
        temp = add(temp, rec(x + 1, y, (s | 1), 0));
    }
    if(y < m && B[y] != c) {
        temp = add(temp, rec(x, y + 1, (s | 2), 1));
    }
    return temp;
}

void solve() {
    cin >> A >> B;
    n = A.size();
    m = B.size();

    int c = 0;
    mem(dp, -1);
    FOR(i,0,n-1) {
        FOR(j,0,m-1) {
            c = add(c, rec(i + 1, j, 1, 0));
            c = add(c, rec(i, j + 1, 2, 1));
        }
    }
    cout << c << endl;
}

int main() {
    ios::sync_with_stdio(false);
    //read;
    //write;
    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
#include <bits/stdc++.h>

using namespace std;

#define mem(a,b) memset(a,b,sizeof(a))
#define FOR(i,j,k) for(int i=j;i<=k;i++)
#define read        freopen("in.txt", "r", stdin)
#define write       freopen("out.txt", "w", stdout)
#define pf          printf
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define PB push_back
#define F first
#define S second
#define MP make_pair

typedef long long ll;
typedef pair<int,int> pii;

const int M = 998244353 ;
const int MAX = (1e3) + 10;

mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());

int n, m, dp[MAX][MAX][4][2];
string A, B;

int add(int x,int y) {
    int r = x + y;
    return (r >= M) ? (r - M) : r;
}

int rec(int x,int y,int s,int t) {
    // cout << x << " " << y << ' ' << s << " " << t << endl;
    int &temp = dp[x][y][s][t];
    if(temp != -1) return temp;

    temp = s == 3 ? 1 : 0;
    char c = t == 0 ? A[x - 1] : B[y - 1];

    if(x < n && A[x] != c) {
        temp = add(temp, rec(x + 1, y, (s | 1), 0));
    }
    if(y < m && B[y] != c) {
        temp = add(temp, rec(x, y + 1, (s | 2), 1));
    }
    return temp;
}

void solve() {
    cin >> A >> B;
    n = A.size();
    m = B.size();

    int c = 0;
    mem(dp, -1);
    FOR(i,0,n-1) {
        FOR(j,0,m-1) {
            c = add(c, rec(i + 1, j, 1, 0));
            c = add(c, rec(i, j + 1, 2, 1));
        }
    }
    cout << c << endl;
}

int main() {
    ios::sync_with_stdio(false);
    //read;
    //write;
    solve();
    return 0;
}


```