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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Scanner;

public final class Main {

    private static final int MOD = 998244353;

    public static void main(String[] args) {
        final Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        final int a = in.nextInt();
        final int b = in.nextInt();
        final int c = in.nextInt();
        final int d = in.nextInt();
        System.out.println(dfs(a, b, c, d, new Integer[c + 1][d + 1]));
    }

    private static long dfs(int a, int b, int c, int d, Integer[][] dp) {
        if (a == c && b == d) {
            return 1;
        }
        if (dp[a][b] != null) {
            return dp[a][b];
        }
        int res = 0;
        if (a < c) {
            res += Math.floorMod(b * dfs(a + 1, b, c, d, dp), MOD);
        }
        if (b < d) {
            res += Math.floorMod(a * dfs(a, b + 1, c, d, dp), MOD);
        }
        if (a < c && b < d) {
            res -= Math.floorMod(a * b * dfs(a + 1, b + 1, c, d, dp), MOD);
        }
        return dp[a][b] = Math.floorMod(res, MOD);
    }
}