//#pragma GCC optimize("Ofast")
//#pragma GCC optimization("unroll-loops, no-stack-protector")
//#pragma GCC target("avx,avx2,fma")

//#include <pair>

#include <vector>
#include<algorithm>
#include <map>
#include <iostream>
#include <string>
#include <queue>
#include <set>

using namespace std;

int M = 998244353;

int addMod(int a, int b) {
    return (a + b) % M;
}

int subMod(int a, int b) {
    return (a - b) % M;
}

void solveTestCase() {
    string a; cin >> a;
    string b; cin >> b;


    vector<vector<vector<int>>> dp(1001, vector<vector<int>>(1001, vector<int>(2, 0)));;
    
    for (int i = 0; i <= a.length(); i++) {
        for (int j = 0; j <= b.length(); j++) {
            if (i < a.length()) dp[i + 1][j][0]++;
            if (j < b.length()) dp[i][j + 1][1]++;
        }
    }
    
    long long r = 0LL;
    int current = 1;
    for (int i = 1; i < a.length(); i++)
    {
        if (a[i] == a[i - 1])
        {
            r = subMod(r, current * (current + 1) / 2 * b.length());
            current = 1;
        }
        else
            current++;
    }
    r = subMod(r, (current * (current + 1) / 2) * b.length());

    current = 1;
    for (int i = 1; i < b.length(); i++)
    {
        if (b[i] == b[i - 1])
        {
            r = subMod(r, current * (current + 1) / 2 * a.length());
            current = 1;
        }
        else
            current++;
    }
    r = subMod(r, (current * (current + 1) / 2) * a.length());

    for (int i = 0; i <= a.length(); i++) {
        for (int j = 0; j <= b.length(); j++) {
            if (i > 0 && i < a.length() && a[i - 1] != a[i])
				dp[i + 1][j][0] = addMod(dp[i + 1][j][0], dp[i][j][0]);
            if (i > 0 && j < b.length() && a[i - 1] != b[j])
                dp[i][j + 1][1] = addMod(dp[i][j + 1][1], dp[i][j][0]);

			if (j > 0 && i < a.length() && b[j - 1] != a[i])
				dp[i + 1][j][0] = addMod(dp[i + 1][j][0], dp[i][j][1]);
            if (j > 0 && j < b.length() && b[j - 1] != b[j])
                dp[i][j + 1][1] = addMod(dp[i][j + 1][1], dp[i][j][1]);
        }
    }

    for (int i = 1; i <= a.length(); i++) {
        for (int j = 1; j <= b.length(); j++) {
            r = addMod(r, dp[i][j][0]);
            r = addMod(r, dp[i][j][1]);
        }
    }
    
    cout << r << endl;
}

int main() {
    solveTestCase();
    return 0;
}

### 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
//#pragma GCC optimize("Ofast")
//#pragma GCC optimization("unroll-loops, no-stack-protector")
//#pragma GCC target("avx,avx2,fma")

//#include <pair>

#include <vector>
#include<algorithm>
#include <map>
#include <iostream>
#include <string>
#include <queue>
#include <set>

using namespace std;

int M = 998244353;

int addMod(int a, int b) {
    return (a + b) % M;
}

int subMod(int a, int b) {
    return (a - b) % M;
}

void solveTestCase() {
    string a; cin >> a;
    string b; cin >> b;


    vector<vector<vector<int>>> dp(1001, vector<vector<int>>(1001, vector<int>(2, 0)));;
    
    for (int i = 0; i <= a.length(); i++) {
        for (int j = 0; j <= b.length(); j++) {
            if (i < a.length()) dp[i + 1][j][0]++;
            if (j < b.length()) dp[i][j + 1][1]++;
        }
    }
    
    long long r = 0LL;
    int current = 1;
    for (int i = 1; i < a.length(); i++)
    {
        if (a[i] == a[i - 1])
        {
            r = subMod(r, current * (current + 1) / 2 * b.length());
            current = 1;
        }
        else
            current++;
    }
    r = subMod(r, (current * (current + 1) / 2) * b.length());

    current = 1;
    for (int i = 1; i < b.length(); i++)
    {
        if (b[i] == b[i - 1])
        {
            r = subMod(r, current * (current + 1) / 2 * a.length());
            current = 1;
        }
        else
            current++;
    }
    r = subMod(r, (current * (current + 1) / 2) * a.length());

    for (int i = 0; i <= a.length(); i++) {
        for (int j = 0; j <= b.length(); j++) {
            if (i > 0 && i < a.length() && a[i - 1] != a[i])
				dp[i + 1][j][0] = addMod(dp[i + 1][j][0], dp[i][j][0]);
            if (i > 0 && j < b.length() && a[i - 1] != b[j])
                dp[i][j + 1][1] = addMod(dp[i][j + 1][1], dp[i][j][0]);

			if (j > 0 && i < a.length() && b[j - 1] != a[i])
				dp[i + 1][j][0] = addMod(dp[i + 1][j][0], dp[i][j][1]);
            if (j > 0 && j < b.length() && b[j - 1] != b[j])
                dp[i][j + 1][1] = addMod(dp[i][j + 1][1], dp[i][j][1]);
        }
    }

    for (int i = 1; i <= a.length(); i++) {
        for (int j = 1; j <= b.length(); j++) {
            r = addMod(r, dp[i][j][0]);
            r = addMod(r, dp[i][j][1]);
        }
    }
    
    cout << r << endl;
}

int main() {
    solveTestCase();
    return 0;
}

```