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<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
int A, B, n, m;
long long D[3010][3010], Mod = 998244353, po1[3010], po2[3010];
long long Pow(long long a, long long b){
    long long r=1;
    while(b){
        if(b&1)r=r*a%Mod;
        b>>=1;a=a*a%Mod;
    }
    return r;
}
int main(){
    int i, j;
    scanf("%d%d%d%d",&A,&B,&n,&m);
    D[A][B]=1;
    for(i=A;i<=n;i++){
        long long t = Pow(i-1,Mod-2);
        po1[0]=po2[0]=1;
        for(j=1;j<=m;j++){
            po1[j]=po1[j-1]*(i-1)%Mod;
            po2[j]=po2[j-1]*t%Mod;
        }
        long long s=0;
        for(j=B;j<=m;j++){
            if(i==A&&j==B)continue;
            D[i][j]=(D[i-1][j]*j+s*po1[j])%Mod;
            s=(s+D[i-1][j]*po2[j])%Mod;
        }
    }
    long long res=0;
    for(i=B;i<=m;i++){
        res+=D[n][i]*Pow(n,m-i)%Mod;
    }
    printf("%lld\n",res%Mod);
}