#include <bits/stdc++.h>
//#define int long long
//#define INT_MAX INT64_MAX
//#define INT_MIN INT64_MIN
#define ll long long
#define PI 3.14159265358979323846
#define CLK clock_t clk = clock();//Start of main
#define OCLK cerr << "Time (in ms): " << (double)(clock() - clk) * 1000.0 / CLOCKS_PER_SEC << '\n';//End of main
#define MOD 998244353
/* DEFINE cout */
#define pln cout << "===============================\n"
#define on cout<<"\n"
#define os cout<<" "
#define o2(a,b) cout<<a<<" "<<b
#define o(a) cout << a
#define bitcount __builtin_popcountll
/* DEFINE cout */
#define endl "\n"
#define fastio ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);//fastio
using namespace std;
string x, y;
int n, m;
int dp[1001][1001][3][2][2];
int fun(int i, int j, int last, bool a, bool b){
    if(i == n && j == m){
        if(a&&b){
            return 1;
        }
        return 0;
    }
    if(dp[i][j][last+1][a][b] != -1){
        return dp[i][j][last+1][a][b];
    }
    ll tr = 0;
    if(a&&b){
        tr++;
    }
    int prev = -1;
    if(last == 0 && i-1>=0){
        prev = x[i-1] - 'a';
    }else if(last == 1 && j-1>=0){
        prev = y[j-1] - 'a';
    }
    if(n > i && x[i] - 'a' != prev){
        tr += fun(i+1, j, 0, true, b);
    }
    if(m > j && y[j] - 'a' != prev){
        tr += fun(i, j+1, 1, a, true);
        
    }
    dp[i][j][last+1][a][b] = tr%MOD;
    return tr%MOD;
    
}
void solve(){
    memset(dp, -1, sizeof(dp));
    cin >> x >> y;
    n = x.length();
    m = y.length();
    ll ans = 0;
    for(int i=0; n>i; i++){
        for(int j=0; m>j; j++){
            ans += fun(i, j, -1, false, false);
            ans %= MOD;
        }
    }
    cout << ans << endl;
    
    
}
signed main(){
    fastio;
    int t = 1;
//    cin >> t;
    while(t--){
        solve();
    }
}






### Prompt

Please provide a CPP coded solution 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 int long long
//#define INT_MAX INT64_MAX
//#define INT_MIN INT64_MIN
#define ll long long
#define PI 3.14159265358979323846
#define CLK clock_t clk = clock();//Start of main
#define OCLK cerr << "Time (in ms): " << (double)(clock() - clk) * 1000.0 / CLOCKS_PER_SEC << '\n';//End of main
#define MOD 998244353
/* DEFINE cout */
#define pln cout << "===============================\n"
#define on cout<<"\n"
#define os cout<<" "
#define o2(a,b) cout<<a<<" "<<b
#define o(a) cout << a
#define bitcount __builtin_popcountll
/* DEFINE cout */
#define endl "\n"
#define fastio ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);//fastio
using namespace std;
string x, y;
int n, m;
int dp[1001][1001][3][2][2];
int fun(int i, int j, int last, bool a, bool b){
    if(i == n && j == m){
        if(a&&b){
            return 1;
        }
        return 0;
    }
    if(dp[i][j][last+1][a][b] != -1){
        return dp[i][j][last+1][a][b];
    }
    ll tr = 0;
    if(a&&b){
        tr++;
    }
    int prev = -1;
    if(last == 0 && i-1>=0){
        prev = x[i-1] - 'a';
    }else if(last == 1 && j-1>=0){
        prev = y[j-1] - 'a';
    }
    if(n > i && x[i] - 'a' != prev){
        tr += fun(i+1, j, 0, true, b);
    }
    if(m > j && y[j] - 'a' != prev){
        tr += fun(i, j+1, 1, a, true);
        
    }
    dp[i][j][last+1][a][b] = tr%MOD;
    return tr%MOD;
    
}
void solve(){
    memset(dp, -1, sizeof(dp));
    cin >> x >> y;
    n = x.length();
    m = y.length();
    ll ans = 0;
    for(int i=0; n>i; i++){
        for(int j=0; m>j; j++){
            ans += fun(i, j, -1, false, false);
            ans %= MOD;
        }
    }
    cout << ans << endl;
    
    
}
signed main(){
    fastio;
    int t = 1;
//    cin >> t;
    while(t--){
        solve();
    }
}






```