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: Accepted)

#include<iostream>
using namespace std;
typedef long long LL;
const LL mod=998244353;
LL dp[3100][3100];

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;i<=c;i++)
        for(int j=b;j<=d;j++)
        {
            if(dp[i][j]!=0)    continue;
            else
            {
                dp[i][j]=(dp[i-1][j]*j%mod+dp[i][j-1]*i%mod-(i-1)*(j-1)*dp[i-1][j-1]%mod+10*mod)%mod;
            }
        }

        cout<<dp[c][d]<<endl;
}