#include <bits/stdc++.h>
using namespace std;

#ifdef LOCAL
#define DEBUG(...) debug(#__VA_ARGS__, __VA_ARGS__)
#else
#define DEBUG(...) 6
#endif

template<typename T, typename S> ostream& operator << (ostream &os, const pair<T, S> &p) {return os << "(" << p.first << ", " << p.second << ")";}
template<typename C, typename T = decay<decltype(*begin(declval<C>()))>, typename enable_if<!is_same<C, string>::value>::type* = nullptr>
ostream& operator << (ostream &os, const C &c) {bool f = true; os << "["; for (const auto &x : c) {if (!f) os << ", "; f = false; os << x;} return os << "]";}
template<typename T> void debug(string s, T x) {cerr << s << " = " << x << "\n";}
template <typename T, typename... Args> void debug(string s, T x, Args... args) {for (int i=0, b=0; i<(int)s.size(); i++) if (s[i] == '(' || s[i] == '{') b++; else
if (s[i] == ')' || s[i] == '}') b--; else if (s[i] == ',' && b == 0) {cerr << s.substr(0, i) << " = " << x << " | "; debug(s.substr(s.find_first_not_of(' ', i + 1)), args...); break;}}

template<int MOD>
struct ModInt {
    long long v;
    ModInt(long long _v = 0) {v = (-MOD < _v && _v < MOD) ? _v : _v % MOD; if (v < 0) v += MOD;}
    ModInt& operator += (const ModInt &other) {v += other.v; if (v >= MOD) v -= MOD; return *this;}
    ModInt& operator -= (const ModInt &other) {v -= other.v; if (v < 0) v += MOD; return *this;}
    ModInt& operator *= (const ModInt &other) {v = v * other.v % MOD; return *this;}
    ModInt& operator /= (const ModInt &other) {return *this *= inverse(other);}
    bool operator == (const ModInt &other) const {return v == other.v;}
    bool operator != (const ModInt &other) const {return v != other.v;}
    friend ModInt operator + (ModInt a, const ModInt &b) {return a += b;}
    friend ModInt operator - (ModInt a, const ModInt &b) {return a -= b;}
    friend ModInt operator * (ModInt a, const ModInt &b) {return a *= b;}
    friend ModInt operator / (ModInt a, const ModInt &b) {return a /= b;}
    friend ModInt operator - (const ModInt &a) {return 0 - a;}
    friend ModInt power(ModInt a, long long b) {ModInt ret(1); while (b > 0) {if (b & 1) ret *= a; a *= a; b >>= 1;} return ret;}
    friend ModInt inverse(ModInt a) {return power(a, MOD - 2);}
    friend ostream& operator << (ostream &os, const ModInt &m) {return os << m.v;}
};
typedef ModInt<998244353> M;

// position in x, last character, have we taken from x, have we taken from y
M dp[1005][26][2][2], ndp[1005][26][2][2];

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

    string x, y;
    cin >> x >> y;
    int n = (int) x.size(), m = (int) y.size();

    M ret = 0;
    for (int k=0; k<26; k++)
        dp[n][k][1][1] = 1;
    for (int i=n-1; i>=0; i--)
        for (int k=0; k<26; k++) {
            dp[i][k][1][1] = 1;
            for (int a=0; a<2; a++)
                for (int b=0; b<2; b++) {
                    bool f = a == 0 && b == 0;
                    if ((f && k == 0) || (!f && x[i] - 'a' != k))
                        dp[i][k][a][b] += dp[i+1][x[i]-'a'][1][b];
                }
        }
    for (int j=m-1; j>=0; j--) {
        for (int i=0; i<=n; i++)
            for (int k=0; k<26; k++)
                for (int a=0; a<2; a++)
                    for (int b=0; b<2; b++)
                        ndp[i][k][a][b] = 0;
        for (int i=n; i>=0; i--)
            for (int k=0; k<26; k++)
                for (int a=0; a<2; a++)
                    for (int b=0; b<2; b++) {
                        bool f = a == 0 && b == 0;
                        // take character from x
                        if (i < n && ((f && k == 0) || (!f && x[i] - 'a' != k)))
                            ndp[i][k][a][b] += ndp[i+1][x[i]-'a'][1][b];
                        // take character from y
                        if ((f && k == 0) || (!f && y[j] - 'a' != k))
                            ndp[i][k][a][b] += dp[i][y[j]-'a'][a][1];
                        // terminate string here
                        if (a && b)
                            ndp[i][k][a][b] += 1;
                    }
        memcpy(dp, ndp, sizeof(ndp));
        for (int i=0; i<=n; i++)
            ret += dp[i][0][0][0];
    }
    cout << ret << "\n";

    return 0;
}

### 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
#include <bits/stdc++.h>
using namespace std;

#ifdef LOCAL
#define DEBUG(...) debug(#__VA_ARGS__, __VA_ARGS__)
#else
#define DEBUG(...) 6
#endif

template<typename T, typename S> ostream& operator << (ostream &os, const pair<T, S> &p) {return os << "(" << p.first << ", " << p.second << ")";}
template<typename C, typename T = decay<decltype(*begin(declval<C>()))>, typename enable_if<!is_same<C, string>::value>::type* = nullptr>
ostream& operator << (ostream &os, const C &c) {bool f = true; os << "["; for (const auto &x : c) {if (!f) os << ", "; f = false; os << x;} return os << "]";}
template<typename T> void debug(string s, T x) {cerr << s << " = " << x << "\n";}
template <typename T, typename... Args> void debug(string s, T x, Args... args) {for (int i=0, b=0; i<(int)s.size(); i++) if (s[i] == '(' || s[i] == '{') b++; else
if (s[i] == ')' || s[i] == '}') b--; else if (s[i] == ',' && b == 0) {cerr << s.substr(0, i) << " = " << x << " | "; debug(s.substr(s.find_first_not_of(' ', i + 1)), args...); break;}}

template<int MOD>
struct ModInt {
    long long v;
    ModInt(long long _v = 0) {v = (-MOD < _v && _v < MOD) ? _v : _v % MOD; if (v < 0) v += MOD;}
    ModInt& operator += (const ModInt &other) {v += other.v; if (v >= MOD) v -= MOD; return *this;}
    ModInt& operator -= (const ModInt &other) {v -= other.v; if (v < 0) v += MOD; return *this;}
    ModInt& operator *= (const ModInt &other) {v = v * other.v % MOD; return *this;}
    ModInt& operator /= (const ModInt &other) {return *this *= inverse(other);}
    bool operator == (const ModInt &other) const {return v == other.v;}
    bool operator != (const ModInt &other) const {return v != other.v;}
    friend ModInt operator + (ModInt a, const ModInt &b) {return a += b;}
    friend ModInt operator - (ModInt a, const ModInt &b) {return a -= b;}
    friend ModInt operator * (ModInt a, const ModInt &b) {return a *= b;}
    friend ModInt operator / (ModInt a, const ModInt &b) {return a /= b;}
    friend ModInt operator - (const ModInt &a) {return 0 - a;}
    friend ModInt power(ModInt a, long long b) {ModInt ret(1); while (b > 0) {if (b & 1) ret *= a; a *= a; b >>= 1;} return ret;}
    friend ModInt inverse(ModInt a) {return power(a, MOD - 2);}
    friend ostream& operator << (ostream &os, const ModInt &m) {return os << m.v;}
};
typedef ModInt<998244353> M;

// position in x, last character, have we taken from x, have we taken from y
M dp[1005][26][2][2], ndp[1005][26][2][2];

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

    string x, y;
    cin >> x >> y;
    int n = (int) x.size(), m = (int) y.size();

    M ret = 0;
    for (int k=0; k<26; k++)
        dp[n][k][1][1] = 1;
    for (int i=n-1; i>=0; i--)
        for (int k=0; k<26; k++) {
            dp[i][k][1][1] = 1;
            for (int a=0; a<2; a++)
                for (int b=0; b<2; b++) {
                    bool f = a == 0 && b == 0;
                    if ((f && k == 0) || (!f && x[i] - 'a' != k))
                        dp[i][k][a][b] += dp[i+1][x[i]-'a'][1][b];
                }
        }
    for (int j=m-1; j>=0; j--) {
        for (int i=0; i<=n; i++)
            for (int k=0; k<26; k++)
                for (int a=0; a<2; a++)
                    for (int b=0; b<2; b++)
                        ndp[i][k][a][b] = 0;
        for (int i=n; i>=0; i--)
            for (int k=0; k<26; k++)
                for (int a=0; a<2; a++)
                    for (int b=0; b<2; b++) {
                        bool f = a == 0 && b == 0;
                        // take character from x
                        if (i < n && ((f && k == 0) || (!f && x[i] - 'a' != k)))
                            ndp[i][k][a][b] += ndp[i+1][x[i]-'a'][1][b];
                        // take character from y
                        if ((f && k == 0) || (!f && y[j] - 'a' != k))
                            ndp[i][k][a][b] += dp[i][y[j]-'a'][a][1];
                        // terminate string here
                        if (a && b)
                            ndp[i][k][a][b] += 1;
                    }
        memcpy(dp, ndp, sizeof(ndp));
        for (int i=0; i<=n; i++)
            ret += dp[i][0][0][0];
    }
    cout << ret << "\n";

    return 0;
}

```