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<iostream>
#include<stdlib.h>
#define mod 998244353

using namespace std;

long long dp[3005][3005];

int main(){
    int a,b,c,d;
    cin>>a>>b>>c>>d;
    dp[a][b]=1;
    for(int i=a; i<=c;++i){
        for(int j=b; j<=d;++j){
            if(i!=a||j!=b){
                dp[i][j]=dp[i-1][j]*j%mod;
                dp[i][j]+=(dp[i][j-1]*i)%mod;
                dp[i][j]-=(dp[i-1][j-1])*(i-1)*(j-1)%mod;
            }
        }
    }
    cout<<dp[c][d]<<endl;
    return 0;
}