#include <bits/stdc++.h>

#define fi first
#define se second
#define p_b push_back
#define pll pair<ll,ll>
#define pii pair<int,int>
#define m_p make_pair
#define all(x) x.begin(),x.end()
#define sset ordered_set
#define sqr(x) (x)*(x)
#define pw(x) (1ll << x)
#define sz(x) (int)x.size()
#define fout(x) {cout << x << "\n"; return; }

using namespace std;
typedef long long ll;
typedef long double ld;
const int N = 1010;
const int M = 1e7;
const ll inf = 1e18;
const int mod = 998244353;

template <typename T> void vout(T s){cout << s << endl;exit(0);}

int dp[N][N][2][4];

int sum(int a, int b){
    a += b;
    if(a >= mod)a -= mod;
    return a;
}

int main(){
    ios_base :: sync_with_stdio(0);
    cin.tie(0);

    string x, y;
    cin >> x >> y;

    int n = sz(x), m = sz(y);

    for(int i = 0; i < n; i++){
        for(int j = 0; j < m; j++){
            dp[i + 1][j][0][1]++;
            dp[i][j + 1][1][2]++;
        }
    }

    char c;

    for(int i = 0; i <= n; i++){
        for(int j = 0; j <= m; j++)
            for(int i1 = 0; i1 < 4; i1++){
                if(i != 0){
                    c = x[i - 1];
                    if(i < n){
                        if(c != x[i]){
                            dp[i + 1][j][0][i1 | 1] = sum(dp[i + 1][j][0][i1 | 1], dp[i][j][0][i1]);
                        }
                    }
                    if(j < m){
                        if(c != y[j]){
                            dp[i][j + 1][1][i1 | 2] = sum(dp[i][j + 1][1][i1 | 2], dp[i][j][0][i1]);
                        }
                    }
                }
                if(j != 0){
                    c = y[j - 1];
                    if(i < n){
                        if(c != x[i]){
                            dp[i + 1][j][0][i1 | 1] = sum(dp[i + 1][j][0][i1 | 1], dp[i][j][1][i1]);
                        }
                    }
                    if(j < m){
                        if(c != y[j]){
                            dp[i][j + 1][1][i1 | 2] = sum(dp[i][j + 1][1][i1 | 2], dp[i][j][1][i1]);
                        }
                    }
                }
            }
    }

    int ans = 0;

    for(int i = 0; i <= n + 1; i++)
        for(int j = 0; j <= m + 1; j++){
            ans = sum(ans, dp[i][j][0][3]);
            ans = sum(ans, dp[i][j][1][3]);
        }

    cout << ans << "\n";

    return 0;
}


### Prompt

Develop a solution in cpp to the problem described below:

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 fi first
#define se second
#define p_b push_back
#define pll pair<ll,ll>
#define pii pair<int,int>
#define m_p make_pair
#define all(x) x.begin(),x.end()
#define sset ordered_set
#define sqr(x) (x)*(x)
#define pw(x) (1ll << x)
#define sz(x) (int)x.size()
#define fout(x) {cout << x << "\n"; return; }

using namespace std;
typedef long long ll;
typedef long double ld;
const int N = 1010;
const int M = 1e7;
const ll inf = 1e18;
const int mod = 998244353;

template <typename T> void vout(T s){cout << s << endl;exit(0);}

int dp[N][N][2][4];

int sum(int a, int b){
    a += b;
    if(a >= mod)a -= mod;
    return a;
}

int main(){
    ios_base :: sync_with_stdio(0);
    cin.tie(0);

    string x, y;
    cin >> x >> y;

    int n = sz(x), m = sz(y);

    for(int i = 0; i < n; i++){
        for(int j = 0; j < m; j++){
            dp[i + 1][j][0][1]++;
            dp[i][j + 1][1][2]++;
        }
    }

    char c;

    for(int i = 0; i <= n; i++){
        for(int j = 0; j <= m; j++)
            for(int i1 = 0; i1 < 4; i1++){
                if(i != 0){
                    c = x[i - 1];
                    if(i < n){
                        if(c != x[i]){
                            dp[i + 1][j][0][i1 | 1] = sum(dp[i + 1][j][0][i1 | 1], dp[i][j][0][i1]);
                        }
                    }
                    if(j < m){
                        if(c != y[j]){
                            dp[i][j + 1][1][i1 | 2] = sum(dp[i][j + 1][1][i1 | 2], dp[i][j][0][i1]);
                        }
                    }
                }
                if(j != 0){
                    c = y[j - 1];
                    if(i < n){
                        if(c != x[i]){
                            dp[i + 1][j][0][i1 | 1] = sum(dp[i + 1][j][0][i1 | 1], dp[i][j][1][i1]);
                        }
                    }
                    if(j < m){
                        if(c != y[j]){
                            dp[i][j + 1][1][i1 | 2] = sum(dp[i][j + 1][1][i1 | 2], dp[i][j][1][i1]);
                        }
                    }
                }
            }
    }

    int ans = 0;

    for(int i = 0; i <= n + 1; i++)
        for(int j = 0; j <= m + 1; j++){
            ans = sum(ans, dp[i][j][0][3]);
            ans = sum(ans, dp[i][j][1][3]);
        }

    cout << ans << "\n";

    return 0;
}


```