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:
"""

ありうるものは？
差の分だけ塗れる

ｄｐ？
dp[i][j] = dp[i-1][j] * j + dp[i][j-1] * i
i,jの中に 丁度i-A,j-B個ある置き方は全部できる

dp[i][j] = dp[i-1][j] * j + dp[i][j-1] * i - dp[i-1][j-1] * (i-1) * (j-1)

i,j置いた時に
i+j個の黒が入っているはず
→ i-1,jにi+j-1こはいってる * j + i-1,jにi+j-1こはいってる* i - どっちも満たしてる

"""

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

dp = [[0] * (D+1) for i in range(C+1)]

dp[A][B] = 1
mod = 998244353

for i in range(A,C+1):

    for j in range(B,D+1):

        if i == A and j == B:
            continue

        dp[i][j] = dp[i-1][j] * j + dp[i][j-1] * i - dp[i-1][j-1] * (i-1) * (j-1)
        dp[i][j] %= mod

print (dp[C][D]%mod)