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 md=998244353;
int a,b,c,d,i,j,k,cur;
long long f[3003][3003][2];
int main() {
  scanf("%d%d%d%d",&a,&b,&c,&d);
  f[a][b][1]=1;
  for (i=a; i<=c; i++) for (j=b; j<=d; j++) for (k=0; k<2; k++) if (f[i][j][k]) {
    f[i+1][j][0]=(f[i+1][j][0]+f[i][j][k]*j)%md;
    cur=(k==1)?i:1;
    f[i][j+1][1]=(f[i][j+1][1]+f[i][j][k]*cur)%md;
  }
  printf("%d\n",int((f[c][d][0]+f[c][d][1])%md));
  return 0;
}