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 = [int(x) for x in input().split()]
dp = [[0 for i in range(d + 1)] for j in range(c + 1)]
for i in range(a+1):
    for j in range(b+1):
        dp[i][j]=1
for i in range(a, c + 1):
    for j in range(b, d + 1):
        if i == a and j == b:
            continue
        if i - 1 >= a:
            dp[i][j] += (dp[i - 1][j] * j)
        if j - 1 >= b:
            dp[i][j] += (dp[i][j - 1] * i)
        if (i - 1) >= a and (j - 1) >= b:
            dp[i][j] -= dp[i-1][j-1]*(i-1)*(j-1)
        dp[i][j] %= 998244353

print(dp[c][d])