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 for _ in range(c - a + 1)] for _ in range(d - b + 1)]
dp[0][0] = 1
for j in range(c - a + 1):
    for i in range(d - b + 1):
        if i == j == 0:
            dp[i][j] = 1
        elif i == 0:
            dp[i][j] = dp[i][j - 1] * b % mod
        elif j == 0:
            dp[i][j] = dp[i - 1][j] * a % mod
        else:
            tmp1 = (j + a - 1) * dp[i - 1][j] % mod
            tmp2 = (i + b - 1) * dp[i][j - 1] % mod
            tmp3 = -dp[i - 1][j - 1] * (i + b - 1) * (j + a - 1) % mod
            tmp4 = (dp[i - 1][j] + dp[i][j - 1]) % mod
            dp[i][j] = (tmp1 + tmp2 + tmp3 + tmp4) % mod
print(dp[-1][-1])