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.PrintWriter;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class Main {
    long MOD = 998244353L;
    long[][] dp = new long[3001][3001];
    void solve(Scanner sc, PrintWriter pw) {
        int A = sc.nextInt();
        int B = sc.nextInt();
        int C = sc.nextInt();
        int D = sc.nextInt();
        dp[A][B] = 1;
        for (int c = A; c <= C; c++) {
            for (int d = B; d <= D; d++) {
                if (c == A && d == B) {
                    continue;
                }
                long ans = 0;
                if (c > A) {
                    long part = dp[c - 1][d];
                    ans = (part * d);
                }
                if (d > B) {
                    long part = dp[c][d - 1];
                    ans = (ans + (part * c));
                }
                if (c > A && d > B) {
                    ans = (ans - (dp[c - 1][d - 1] * (c - 1) * (d - 1)));
                }
                dp[c][d] = ((ans %MOD) + MOD) % MOD;
            }
        }
        pw.println(dp[C][D]);
    }

    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        new Main().solve(sc, pw);
        pw.flush();
        pw.close();
        sc.close();
    }
}