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>
using namespace std;

int a, b, c, d;
const int MOD = 998244353;
int dp[3055][3055];

int main() {
    cin >> a >> b >> c >> d;

    dp[a][b] = 1;

    for (int i = a; i <= c; i++) {
        for (int j = b; j <= d; j++) {
            if (i == a && j == b) continue;

            dp[i][j] = (1ll * dp[i - 1][j] * j % MOD + 1ll * dp[i][j - 1] * i) % MOD;
         //   dp[i][j] = (1ll * dp[i][j] - dp[i - 1][j - 1] + MOD) % MOD;
            dp[i][j] = (1ll * dp[i][j] - 1ll * (i - 1) * (j - 1) % MOD * dp[i - 1][j - 1] % MOD + MOD) % MOD;

        }
    }

    cout << dp[c][d] << endl;

    return 0;
}