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;

int main() {
  int a, b, c, d;
  cin >> a >> b >> c >> d;
  long long mod = 998244353;
  vector<vector<long long>> dp(c+1, vector<long long>(d+1));
  
  for (int i = a; i < c+1; i++){
    for (int j = b; j < d+1; j++){
      if (i == a && j == b){
        dp[i][j] = 1;
      }
      else if (i == a){
        dp[i][j] = dp[i][j-1]*i%mod;
      }
      else if (j == b){
        dp[i][j] = dp[i-1][j]*j%mod;
      }
      else{
        dp[i][j] = dp[i-1][j]*j%mod + dp[i][j-1]*i%mod + mod - dp[i-1][j-1]*(i-1)*(j-1)%mod;
        dp[i][j] %= mod;
      }
    }
  }
  
  cout << dp[c][d] << endl;
}