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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 3000 + 10, mod = 998244353;
ll f[N][N][2];
void add(ll &x,ll y){
    (x += y) %= mod;
}
int main(){
    ios::sync_with_stdio(0);
    int a,b,c,d; cin>>a>>b>>c>>d;
    f[a][b][0] = 1;
    for(int i=a;i<=c;i++){
        for(int j=b;j<=d;j++){
            add(f[i+1][j][0],f[i][j][0]*j);
            add(f[i][j+1][1],f[i][j][0]*i);
            add(f[i+1][j][0],f[i][j][1]);
            add(f[i][j+1][1],f[i][j][1]*i);
        }
    }
    ll ans = 0;
    add(ans,f[c][d][0]);
    add(ans,f[c][d][1]);
    cout<<ans<<endl;
}