#include <bits/stdc++.h>
using i64 = long long;
constexpr int N = 1000, P = 998244353;
struct Modular {
    int x;
    friend Modular operator+(Modular a, Modular const b) {
        return a += b;
    }
    friend Modular operator-(Modular a, Modular const b) {
        return a -= b;
    }
    friend Modular operator*(Modular a, Modular const b) {
        return a *= b;
    }
    friend Modular operator/(Modular a, Modular const b) { 
        return a /= b;
    }
    Modular(int64_t v = 0) {
        x = v % P;
        if (x < 0) x += P;
    }
    Modular& operator+=(Modular const& b) {
        x += b.x;
        if (x >= P) x -= P;
        return *this;
    }
    Modular& operator-=(Modular const& b) {
        x -= b.x;
        if (x < 0) x += P;
        return *this;
    }
    Modular& operator*=(Modular const& b) {
        x = (i64) x * b.x % P;
        return *this;
    }
    Modular& operator/=(Modular const& b) {
        return *this *= inverse(b);
    }
    friend Modular power(Modular a, i64 e) {
        Modular res = 1;
        for (; e; e >>= 1, a *= a)
            if (e & 1) res *= a;
        return res;
    }
    friend Modular inverse(Modular a) {
        return power(a, P - 2);
    }
    friend bool operator==(Modular const& a, Modular const& b) {
        return a.x == b.x;
    }
    friend bool operator!=(Modular const& a, Modular const& b) {
        return a.x != b.x;
    }
    friend std::ostream& operator<<(std::ostream& os, Modular const& a) {
        return os << a.x;
    }
};
using Mint = Modular;
Mint dp[N][N][2], dps[N], dpt[N];
int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::string s, t;
    std::cin >> s >> t;
    int n = s.length(), m = t.length();
    dps[0] = dpt[0] = 1;
    for (int i = 1; i < n; i++) {
        dps[i] = (s[i] != s[i - 1] ? 1 + dps[i - 1] : 1);
    }
    for (int i = 1; i < m; i++) {
        dpt[i] = (t[i] != t[i - 1] ? 1 + dpt[i - 1] : 1);
    }
    Mint ans = 0;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            if (i > 0 && s[i] != s[i - 1])
                dp[i][j][0] += dp[i - 1][j][0];
            if (s[i] != t[j]) {
                dp[i][j][0] += dpt[j];
                if (i > 0)
                    dp[i][j][0] += dp[i - 1][j][1];
            }
            if (j > 0 && t[j] != t[j - 1])
                dp[i][j][1] += dp[i][j - 1][1];
            if (t[j] != s[i]) {
                dp[i][j][1] += dps[i];
                if (j > 0)
                    dp[i][j][1] += dp[i][j - 1][0];
            }
            ans += dp[i][j][0] + dp[i][j][1];
        }
    }
    std::cout << ans << '\n';
}

### Prompt

Please formulate 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
#include <bits/stdc++.h>
using i64 = long long;
constexpr int N = 1000, P = 998244353;
struct Modular {
    int x;
    friend Modular operator+(Modular a, Modular const b) {
        return a += b;
    }
    friend Modular operator-(Modular a, Modular const b) {
        return a -= b;
    }
    friend Modular operator*(Modular a, Modular const b) {
        return a *= b;
    }
    friend Modular operator/(Modular a, Modular const b) { 
        return a /= b;
    }
    Modular(int64_t v = 0) {
        x = v % P;
        if (x < 0) x += P;
    }
    Modular& operator+=(Modular const& b) {
        x += b.x;
        if (x >= P) x -= P;
        return *this;
    }
    Modular& operator-=(Modular const& b) {
        x -= b.x;
        if (x < 0) x += P;
        return *this;
    }
    Modular& operator*=(Modular const& b) {
        x = (i64) x * b.x % P;
        return *this;
    }
    Modular& operator/=(Modular const& b) {
        return *this *= inverse(b);
    }
    friend Modular power(Modular a, i64 e) {
        Modular res = 1;
        for (; e; e >>= 1, a *= a)
            if (e & 1) res *= a;
        return res;
    }
    friend Modular inverse(Modular a) {
        return power(a, P - 2);
    }
    friend bool operator==(Modular const& a, Modular const& b) {
        return a.x == b.x;
    }
    friend bool operator!=(Modular const& a, Modular const& b) {
        return a.x != b.x;
    }
    friend std::ostream& operator<<(std::ostream& os, Modular const& a) {
        return os << a.x;
    }
};
using Mint = Modular;
Mint dp[N][N][2], dps[N], dpt[N];
int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::string s, t;
    std::cin >> s >> t;
    int n = s.length(), m = t.length();
    dps[0] = dpt[0] = 1;
    for (int i = 1; i < n; i++) {
        dps[i] = (s[i] != s[i - 1] ? 1 + dps[i - 1] : 1);
    }
    for (int i = 1; i < m; i++) {
        dpt[i] = (t[i] != t[i - 1] ? 1 + dpt[i - 1] : 1);
    }
    Mint ans = 0;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            if (i > 0 && s[i] != s[i - 1])
                dp[i][j][0] += dp[i - 1][j][0];
            if (s[i] != t[j]) {
                dp[i][j][0] += dpt[j];
                if (i > 0)
                    dp[i][j][0] += dp[i - 1][j][1];
            }
            if (j > 0 && t[j] != t[j - 1])
                dp[i][j][1] += dp[i][j - 1][1];
            if (t[j] != s[i]) {
                dp[i][j][1] += dps[i];
                if (j > 0)
                    dp[i][j][1] += dp[i][j - 1][0];
            }
            ans += dp[i][j][0] + dp[i][j][1];
        }
    }
    std::cout << ans << '\n';
}

```