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;

int main(){
  int A, B, C, D;
  cin >> A >> B >> C >> D;
  
  vector<vector<long long>> dp(C + 1, vector<long long>(D + 1, 0));
  dp[A][B] = 1;
  for(int a = A; a <= C; a++){
    for(int b = B; b <= D; b++){
      if(a == A && B == b) continue;
      dp[a][b] = dp[a - 1][b] * b + dp[a][b - 1] * a - dp[a - 1][b - 1] * ((a - 1) * (b - 1));
      dp[a][b] %= mod;
    }
  }
  
  
  cout << (mod + dp[C][D]) % mod << endl;
  
  return 0;
}