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:
#from collections import deque,defaultdict
from sys import stdin
input = stdin.readline
printn = lambda x: print(x,end='')
inn = lambda : int(input())
inl   = lambda: list(map(int, input().split()))
inm   = lambda:      map(int, input().split())
ins = lambda : input().strip()
DBG = True # and False
BIG = 10**18
R = 998244353

def ddprint(x):
  if DBG:
    print(x)

a,b,c,d = inm()
dp = [[0]*(d+1) for i in range(c+1)]
dp[a][b] = 1
for i in range(a+1,c+1):
    dp[i][b] = (dp[i-1][b]*b)%R
for j in range(b+1,d+1):
    dp[a][j] = (dp[a][j-1]*a)%R
for i in range(a+1,c+1):
    for j in range(b+1,d+1):
        dp[i][j] = (dp[i-1][j]*j+dp[i][j-1]*i-dp[i-1][j-1]*(i-1)*(j-1))%R
print(dp[c][d])