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;
 
//#pragma GCC optimize("O3")
//#pragma GCC target("sse4")
 
#define FOR(i, a, b) for(int i = (a); i < (b); ++i)
#define RFOR(i, b, a) for(int i = (b) - 1; i >= (a); --i)
#define FILL(a, value) memset(a, value, sizeof(a))
 
#define SZ(a) (int)((a).size())
#define ALL(a) a.begin(), a.end()
 
typedef long long LL;
typedef pair<int, int> PII;

const int mod = 998244353;
const int MAX = 3007;

int add(int a, int b)
{
	return a + b < mod ? a + b : a + b - mod;
}

int sub(int a, int b)
{
	return a - b >= 0 ? a - b : a - b + mod;
}

int mult(int a, int b)
{
	return (LL)a * b % mod;
}

int dp[MAX][MAX];

int main()
{
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	int a, b, c, d;
	cin >> a >> b >> c >> d;
	FOR(i, a, c + 1)
		FOR(j, b, d + 1)
		{
			if(i == a && j == b)
				dp[a][b] = 1;
			else
				dp[i][j] = sub(add(mult(i, dp[i][j - 1]), mult(j, dp[i - 1][j])), mult(mult(i - 1, j - 1), dp[i - 1][j - 1]));
		}
	cout << dp[c][d] << "\n";
	return 0;
}