Problem: We have a grid with A horizontal rows and B vertical columns, with the squares painted white. On this grid, we will repeatedly apply the following operation:

* Assume that the grid currently has a horizontal rows and b vertical columns. Choose "vertical" or "horizontal".
* If we choose "vertical", insert one row at the top of the grid, resulting in an (a+1) \times b grid.
* If we choose "horizontal", insert one column at the right end of the grid, resulting in an a \times (b+1) grid.
* Then, paint one of the added squares black, and the other squares white.



Assume the grid eventually has C horizontal rows and D vertical columns. Find the number of ways in which the squares can be painted in the end, modulo 998244353.

Constraints

* 1 \leq A \leq C \leq 3000
* 1 \leq B \leq D \leq 3000
* A, B, C, and D are integers.

Input

Input is given from Standard Input in the following format:


A B C D


Output

Print the number of ways in which the squares can be painted in the end, modulo 998244353.

Examples

Input

1 1 2 2


Output

3


Input

2 1 3 4


Output

65


Input

31 41 59 265


Output

387222020

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define REP(i, n) for (int i = 0; i < (n); i++)
using namespace std;
using ll = long long;

const int MOD = 998244353;

int main() {
    cin.tie(0)->sync_with_stdio(false);

    int a, b, c, d;
    cin >> a >> b >> c >> d;

    vector<vector<ll> > dp(c + 1, vector<ll>(d + 1, 0));

    dp[a][b] = 1;

    for (int i = a; i <= c; i++) for (int j = b; j <= d; j++)
        if (!(i == a && j == b))
            dp[i][j] = ((dp[i - 1][j] * j) + (dp[i][j - 1] * i) - (dp[i - 1][j - 1] * (i - 1) * (j - 1)) % MOD + MOD) % MOD;

    cout << dp[c][d] << '\n';
    return 0;
}