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 re register
int f[3010][3010];
const int mod=998244353;
int main(){
  re int a, b, c, d;
  scanf("%d%d%d%d", &a, &b, &c, &d);
  f[a][b] = 1;
  for(re int i=a+1; i<=c; i++){
    re int nw = 0;
    for(re int j=b; j<=d; j++){
      f[i][j] = (nw+1ll * f[i-1][j] * j) % mod;
      nw = (1ll * (nw + f[i-1][j]) * (i-1)) % mod;
    }
  }
  re int ans = 0;
  for(re int j=b; j<=d; j++){
    ans = (1ll * ans * c + f[c][j]) % mod;
  }
  printf("%d\n", ans);
}