Problem: 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.

Examples

Input

1 1 2 2


Output

3


Input

2 1 3 4


Output

65


Input

31 41 59 265


Output

387222020

Solution (in JAVA):

Solution:
import java.util.*;

public class Main {

	static int mod = 998244353;
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int A = sc.nextInt();
		int B = sc.nextInt();
		int C = sc.nextInt();
		int D = sc.nextInt();

		long[][] dp = new long[C+1][D+1];
		dp[A][B] = 1L;
		for (int j=B;j<D;j++) {
			dp[A][j+1] = dp[A][j]*A%mod;
		}
		for (int i=A;i<C;i++) {
			dp[i+1][B] = dp[i][B]*B%mod;
		}
		for (int i=A;i<C;i++) {
			for (int j=B;j<D;j++) {
				dp[i+1][j+1] = (dp[i+1][j]*(i+1)%mod+dp[i][j+1]*(j+1)%mod-dp[i][j]*(long)i%mod*j%mod+mod)%mod;
			}
		}
		System.out.println(dp[C][D]);
	}
}