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;
using ll = long long;
using ld = long double;
template<typename T> using v2 = vector<vector<T>>;
template<typename T> inline v2<T> fill(int r, int c, T t){return v2<T>(r, vector<T>(c, t));}

const int MOD = 998244353;
inline int add(int a, int b){return (a+b)%MOD;}
inline int sub(int a, int b){return ((a-b)%MOD + MOD)%MOD;}
inline int mul(int a, int b){return ((ll)a*b)%MOD;}

int a, b, c, d;
v2<int> dp;
void solve(){
    cin >> a >> b >> c >> d;
    dp = fill(c+1, d+1, -1);
    for(int i = a; i <= c; i++){
        for(int j = b; j <= d; j++){
            if(i==a && j==b) dp[i][j] = 1;
            else if(i==a) dp[i][j] = mul(dp[i][j-1], i);
            else if(j==b) dp[i][j] = mul(dp[i-1][j], j);
            else dp[i][j] = sub(add(mul(dp[i][j-1], i), mul(dp[i-1][j], j)), mul(mul(i-1, j-1), dp[i-1][j-1]));
        }
    }
        
    cout << dp[c][d] << '\n';
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
    return 0;
}