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

dp1=[[0]*(d+1) for _ in range(c+1)]
dp2=[[0]*(d+1) for _ in range(c+1)]

for i in range(b+1,d+1):
  if a==1:
    dp1[a][i]=1 if i==b+1 else 0
    dp2[a][i]=0 if i==b+1 else 1
  else:
    dp1[a][i]=(pow(a-1,i-b-1,mod)*(i-b))%mod
    dp2[a][i]=(pow(a,i-b,mod)-dp1[a][i])%mod
for i in range(a+1,c+1):
  dp1[i][b]=pow(b,i-a,mod)
  dp2[i][b]=0
dp2[a][b]=1
for i in range(a+1,c+1):
  for j in range(b+1,d+1):
    dp1[i][j]=(j*(dp1[i-1][j]+dp2[i-1][j]))%mod
    dp2[i][j]=(dp1[i][j-1]+dp2[i][j-1]*i)%mod
print((dp1[c][d]+dp2[c][d])%mod)

"""
print('dp1')
for x in dp1:
  print(x)
print('dp2')
for x in dp2:
  print(x)
"""