//#define ACM_LOCAL 1
#if 1

#include<bits/stdc++.h>

#ifdef ONLINE_JUDGE
#define endl '\n'
#endif
using namespace std;
typedef long long ll;
const ll mod(998244353);
const ll MAXN(2e7 + 7);
const double eps(1e-6);

inline ll powmod(ll a, ll b) {
    ll res = 1;
    a %= mod;
    assert(b >= 0);
    for (; b; b >>= 1) {
        if (b & 1)res = res * a % mod;
        a = a * a % mod;
    }
    return res % mod;
}

inline ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }

struct GlobalData {
    string s, t;
    int ans;

public:
    void init() {
        ans = 0;
    }

    void input() {
        cin >> s >> t;
    }

    void outputAns() {
        cout << ans << endl;
    }

    void preSolve() {}
} globalData;

inline void preSolve() {
    globalData.preSolve();
}

inline void input() {
    globalData.init();
    globalData.input();
}

inline void output() {
    globalData.outputAns();
}

inline void addMod(int &a, const int &b) {
    a = (1ll*a+b)%mod;
}

inline void solve() {

    int n = globalData.s.size(), m = globalData.t.size();
    vector<vector<vector<int>>> dp(n+1, vector<vector<int>>(m+1, vector<int>(3)));
    dp[0][0][2] = 1;
    for(int i=0; i<=n; i++) {
        for(int j=0; j<=m; j++) {
            dp[i][j][2] = 1;    // 无前缀情况
            // dp[i+1][j][0] = dp[i][j][0] + dp[i][j][1] + dp[i][j][2]
            if(i<n) {
                if (j > 0 && globalData.s[i] != globalData.t[j - 1]) addMod(dp[i + 1][j][0], dp[i][j][1]);
                if (i > 0 && globalData.s[i] != globalData.s[i - 1]) addMod(dp[i + 1][j][0], dp[i][j][0]);
                addMod(dp[i + 1][j][0], dp[i][j][2]);
            }
            if(j<m) {
                if (j > 0 && globalData.t[j] != globalData.t[j - 1]) addMod(dp[i][j + 1][1], dp[i][j][1]);
                if (i > 0 && globalData.t[j] != globalData.s[i - 1]) addMod(dp[i][j + 1][1], dp[i][j][0]);
                addMod(dp[i][j + 1][1], dp[i][j][2]);
            }
        }
    }
    for(int i=1; i<=n; i++) for(int j=1; j<=m; j++) for(int k=0; k<2; k++) addMod(globalData.ans, dp[i][j][k]);
    int val = 0;
    for(int i=0; i<n; i++) {
        if(i>0 && globalData.s[i]!=globalData.s[i-1]) val += m;
        else val = m;
        globalData.ans -= val;
    }
    for(int j=0; j<m; j++) {
        if(j>0 && globalData.t[j]!=globalData.t[j-1]) val += n;
        else val = n;
        globalData.ans -= val;
    }
    globalData.ans = (1ll * globalData.ans + mod)%mod;

}

int main() {

#ifdef ACM_LOCAL
    freopen("./data/9.in", "r", stdin);
    freopen("./data/9.out", "w", stdout);
#endif
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

    preSolve();
    int T = 1;
//    cin >> T;
    while (T--) {
        input();
        solve();
        output();
    }

    return 0;

}

#endif
### Prompt

Your task is to create 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
//#define ACM_LOCAL 1
#if 1

#include<bits/stdc++.h>

#ifdef ONLINE_JUDGE
#define endl '\n'
#endif
using namespace std;
typedef long long ll;
const ll mod(998244353);
const ll MAXN(2e7 + 7);
const double eps(1e-6);

inline ll powmod(ll a, ll b) {
    ll res = 1;
    a %= mod;
    assert(b >= 0);
    for (; b; b >>= 1) {
        if (b & 1)res = res * a % mod;
        a = a * a % mod;
    }
    return res % mod;
}

inline ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }

struct GlobalData {
    string s, t;
    int ans;

public:
    void init() {
        ans = 0;
    }

    void input() {
        cin >> s >> t;
    }

    void outputAns() {
        cout << ans << endl;
    }

    void preSolve() {}
} globalData;

inline void preSolve() {
    globalData.preSolve();
}

inline void input() {
    globalData.init();
    globalData.input();
}

inline void output() {
    globalData.outputAns();
}

inline void addMod(int &a, const int &b) {
    a = (1ll*a+b)%mod;
}

inline void solve() {

    int n = globalData.s.size(), m = globalData.t.size();
    vector<vector<vector<int>>> dp(n+1, vector<vector<int>>(m+1, vector<int>(3)));
    dp[0][0][2] = 1;
    for(int i=0; i<=n; i++) {
        for(int j=0; j<=m; j++) {
            dp[i][j][2] = 1;    // 无前缀情况
            // dp[i+1][j][0] = dp[i][j][0] + dp[i][j][1] + dp[i][j][2]
            if(i<n) {
                if (j > 0 && globalData.s[i] != globalData.t[j - 1]) addMod(dp[i + 1][j][0], dp[i][j][1]);
                if (i > 0 && globalData.s[i] != globalData.s[i - 1]) addMod(dp[i + 1][j][0], dp[i][j][0]);
                addMod(dp[i + 1][j][0], dp[i][j][2]);
            }
            if(j<m) {
                if (j > 0 && globalData.t[j] != globalData.t[j - 1]) addMod(dp[i][j + 1][1], dp[i][j][1]);
                if (i > 0 && globalData.t[j] != globalData.s[i - 1]) addMod(dp[i][j + 1][1], dp[i][j][0]);
                addMod(dp[i][j + 1][1], dp[i][j][2]);
            }
        }
    }
    for(int i=1; i<=n; i++) for(int j=1; j<=m; j++) for(int k=0; k<2; k++) addMod(globalData.ans, dp[i][j][k]);
    int val = 0;
    for(int i=0; i<n; i++) {
        if(i>0 && globalData.s[i]!=globalData.s[i-1]) val += m;
        else val = m;
        globalData.ans -= val;
    }
    for(int j=0; j<m; j++) {
        if(j>0 && globalData.t[j]!=globalData.t[j-1]) val += n;
        else val = n;
        globalData.ans -= val;
    }
    globalData.ans = (1ll * globalData.ans + mod)%mod;

}

int main() {

#ifdef ACM_LOCAL
    freopen("./data/9.in", "r", stdin);
    freopen("./data/9.out", "w", stdout);
#endif
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

    preSolve();
    int T = 1;
//    cin >> T;
    while (T--) {
        input();
        solve();
        output();
    }

    return 0;

}

#endif
```