Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

1 1 2 2

Sample Output 1

3

Any two of the three squares other than the bottom-left square can be painted black.

Sample Input 2

2 1 3 4

Sample Output 2

65

Sample Input 3

31 41 59 265

Sample Output 3

387222020

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

long long mod = 998244353;
long long A,B,C,D;
long long dp[3030][3030];

int main(){
  cin >> A >> B >> C >> D;
  dp[A][B] = 1;
  for (long long i = A; i <= C; i++){
    for (long long j = B; j <= D; j++){
      if ((i > A)||(j > B)){
        dp[i][j]=
          (i * dp[i][j-1]  
          + j * dp[i-1][j] 
          - (i - 1) * (j - 1) * dp[i-1][j-1]) % mod;
      }
    }
  }
  cout << dp[C][D];
}