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 long long MOD=998244353;
long long dp[3009][3009];

int main(){
  int a, b, c, d;
  cin >> a >> b >> c >> d;
  dp[a][b]=1;
  for(int i=a+1;i<=c;i++){
    dp[i][b]=(b*dp[i-1][b])%MOD;
  }
  for(int i=b+1;i<=d;i++){
    dp[a][i]=(a*dp[a][i-1])%MOD;
  }

  for(int i=a+1;i<=c;i++){
    for(int j=b+1;j<=d;j++){
      dp[i][j] = (i*dp[i][j-1])%MOD;
      dp[i][j] = (dp[i][j] + (j*dp[i-1][j])%MOD)%MOD;
      dp[i][j] = (dp[i][j] - ((i-1)*(j-1)*dp[i-1][j-1])%MOD + MOD)%MOD;
    }
  }
  cout << dp[c][d] << endl;
}