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>
#include<iomanip>
#include<cmath>
#include<string>
#include<cstring>
#include<vector>
#include<list>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include<stack>
using namespace std;
typedef long long ll;
#define fi first
#define se second
#define mp make_pair
#define rep(i, n) for(int i=0;i<n;++i)
#define rrep(i, n) for(int i=n;i>=0;--i)
const int inf=1e9+7;
const ll mod=1e9+7;
const ll mod1=998244353;
const ll big=1e18;
const double PI=2*asin(1);

ll DP[3005][3005];

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] = DP[i-1][B]*B%mod1;
  }
  for(int j=B+1;j<=D;++j) {
    DP[A][j] = DP[A][j-1]*A%mod1;
  }
  for(ll i=A+1;i<=C;++i) {
    for(ll j=B+1;j<=D;++j) {
      DP[i][j] = DP[i-1][j]*j%mod1 + DP[i][j-1]*i%mod1 - DP[i-1][j-1]*(i-1)%mod1*(j-1)%mod1;
      DP[i][j] = (DP[i][j]+mod1)%mod1;
    }
  }
  cout<<DP[C][D]<<endl;
}