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:
# coding: utf-8
import sys

sr = lambda: sys.stdin.readline().rstrip()
ir = lambda: int(sr())
lr = lambda: list(map(int, sr().split()))

"""
各マスのdp
下から来たものと左から来たものを足す
その後、重なっているところを引く
"""
MOD = 998244353
A, B, C, D = lr()  # Cは縦
dp = [[0] * (D+1) for _ in range(C+1)]   # 1-indexed
dp[A][B] = 1
for i in range(A, C+1):
    for j in range(B, D+1):
        # iが縦
        if i > A:
            dp[i][j] += dp[i-1][j] * j
        if j > B:
            dp[i][j] += dp[i][j-1] * i
        if i > A and j > B:
            dp[i][j] -= dp[i-1][j-1] * (i-1) * (j-1)
        dp[i][j] %= MOD

answer = dp[C][D]
print(answer%MOD)