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;
const ll MOD=998244353;
ll dp[4001][4001],a,b,c,d;
ll f(ll A,ll B){
  if(A<a||B<b)return 0;
  if(~dp[A][B])return dp[A][B];
  return dp[A][B]=(MOD+f(A-1,B)*B+f(A,B-1)*A-f(A-1,B-1)*(A-1)*(B-1)%MOD)%MOD;
}

int main(){
  cin>>a>>b>>c>>d;
  memset(dp,-1,sizeof(dp));
  dp[a][b]=1;
  cout<<f(c,d)<<endl;
}