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 d[3005][3005],mod = 998244353;
int A,B,C,D;

int main() {
    ios_base::sync_with_stdio(false); cin.tie(0);
    cin >> A >> B >> C >> D;
    d[A][B] = 1;
    for(int i = A;i <= C;i++) {
        for(int j = B;j <= D;j++) {
            if(i == A&&j == B) continue;
            d[i][j] = ((d[i-1][j]*j+d[i][j-1]*i-d[i-1][j-1]*(i-1)*(j-1))%mod+mod)%mod;
        }
    }
    cout << d[C][D];
}