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;
const int mod = 998244353;
#define LL long long
LL f[3010][3010];
int main(){
    int A,B,C,D;
    scanf("%d%d%d%d",&A,&B,&C,&D);
    f[A][B] = 1;
    for (int i=A; i<=C; i++){
        for (int j=B; j<=D; j++){
            if (i == A && j == B) continue;
            else if (i == A) f[i][j] = f[i][j-1]*i%mod;
            else if (j == B) f[i][j] = f[i-1][j]*j%mod;
            else f[i][j] = ((f[i-1][j]*j%mod + f[i][j-1]*i%mod)%mod - f[i-1][j-1]*(i-1)%mod*(j-1)%mod + mod)%mod;
        }
    }
    printf("%lld\n",f[C][D]);
    return 0;
}