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 PYTHON3):

Solution:
import sys

sys.setrecursionlimit(10 ** 8)

input = sys.stdin.readline


def main():
    A, B, C, D = [int(x) for x in input().split()]

    MOD = 998244353

    dp = [[0] * (D - B + 1) for j in range(C - A + 1)]

    for j in range(C - A + 1):
        if j == 0:
            dp[j][0] = 1
        else:
            dp[j][0] = pow(B, j, MOD)
        for i in range(1, D - B + 1):
            if j == 0:
                dp[j][i] = dp[j][i - 1] * (A + j) % MOD
            else:
                dp[j][i] = (dp[j][i - 1] * (A + j) + (B + i) * dp[j - 1][i] - dp[j - 1][i - 1] * (
                            (A + j - 1) * (B + i - 1))) % MOD

    print(dp[-1][-1] % MOD)


if __name__ == '__main__':
    main()