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:
import sys
sys.setrecursionlimit(10**7)

MOD = 998244353
A,B,C,D= map(int,input().split())
#A,B,C,D = 1,1,3000,3000

def dpinit(ps, val=0):
    res = [val for i in [0]*ps[-1]]
    for i in ps[:-1][::-1]:
        res = [res[:] for k in [0]*i]
    return res

memo = dpinit((C+1,D+1))

memo[A][B] = 1
for Y in range(A,C+1):
    memo[Y][B] = pow(B,Y-A,MOD)
for X in range(B,D+1):
    memo[A][X] = pow(A,X-B,MOD)

for y in range(A+1,C+1):
    for x in range(B+1,D+1):
        memo[y][x] = (x*memo[y-1][x] + y*memo[y][x-1] - (x-1)*(y-1)*memo[y-1][x-1]) % MOD

print(memo[C][D])