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:
MOD=998244353
a,b,c,d=map(int,input().split())
dp=[[0 for j in range(d+1)] for i in range(c+1)]
dp[a][b]=1
for i in range(a+1,c+1):
    dp[i][b]=(b*dp[i-1][b])%MOD
for i in range(b+1,d+1):
    dp[a][i]=(a*dp[a][i-1])%MOD
for i in range(a+1,c+1):
    for j in range(b+1,d+1):
      dp[i][j]=(i*dp[i][j-1])%MOD
      dp[i][j]=(dp[i][j]+(j*dp[i-1][j])%MOD)%MOD
      dp[i][j]=(dp[i][j]-((i-1)*(j-1)*dp[i-1][j-1])%MOD)%MOD
print(dp[c][d])