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())
mod = 998244353
dp = [[0]*(d-b+1) for i in range(c-a+1)]
if a == c and b == d:
    print(1)
elif a == c:
    print(pow(a,d-b,mod))
elif b == d:
    print(pow(b,c-a,mod))
else:
    for i in range(c-a):
        p = 0
        for j in range(d-b+1):
            if i == j == 0:
                dp[i][j] = 1
            dp[i+1][j] = (p + dp[i][j]*(b+j))%mod

            p += dp[i][j]
            p %= mod
            p *= a+i
            p %= mod

    ans = 0
    for d in dp[-1]:
        ans *= c
        ans %= mod
        ans += d
        ans %= mod
    print(ans)