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;
typedef long long ll;
ll mod = 998244353;
ll dp[3010][3010];
ll A,B,C,D;
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++) {
            dp[i][j] += dp[i-1][j] * j;
            dp[i][j] %= mod;
            dp[i][j] += dp[i][j-1] * i;
            dp[i][j] %= mod;
            dp[i][j] -= dp[i-1][j-1] * (i - 1) % mod * (j - 1) % mod;
            if (dp[i][j] < 0)
                dp[i][j] += mod * mod;
            dp[i][j] %= mod;
        }
    }
    cout << dp[C][D] << endl;
    return 0;
}