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 y in range(D-B+1)]for x in range(C-A+1)]
dp[0][0]=1
for x in range(C-A):
    dp[x+1][0]=(dp[x][0]*B)%mod
for y in range(D-B):
    dp[0][y+1]=(dp[0][y]*A)%mod
for x in range(C-A):
    for y in range(D-B):
        dp[x+1][y+1]=(dp[x+1][y]*(A+x))%mod
        dp[x+1][y+1]+=(dp[x][y+1]*(B+y))%mod
        dp[x+1][y+1]%=mod
        dp[x+1][y+1]-=(dp[x][y]*(A+x)*(B+y))%mod
        dp[x+1][y+1]%=mod
        dp[x+1][y+1]+=(dp[x+1][y]+dp[x][y+1])%mod
        dp[x+1][y+1]%=mod

print(dp[C-A][D-B])