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:
A,B,C,D=map(int, input().split())

#上と右

p=998244353
dp=[[0]*(D+1) for _ in range(C+1)]



for h in range(A,C+1):
    for w in range(B,D+1):
        if h==A and w==B:
            dp[A][B]=1
            continue

        if h>A and w>B:
            x=dp[h][w-1]*h
            y=dp[h-1][w]*w
            z=dp[h-1][w-1]*(h-1)*(w-1)
            sub=x+y-z
            dp[h][w]=(sub)%p
        elif h==A:
            dp[h][w]=(dp[h][w-1]*h)%p
        else:
            dp[h][w]=(dp[h-1][w]*w)%p


print(dp[C][D])