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
input = sys.stdin.readline

def I(): return int(input())
def MI(): return map(int, input().split())
def LI(): return list(map(int, input().split()))

def main():
    mod=998244353
    A,B,C,D=MI()
    
    if A==C and B==D:
        print(1)
        exit()
    
    F=[[0]*(D+1) for _ in range(C+1)]
    
    def calc(a,b):
        temp=F[a][b-1]*a + F[a-1][b]*b - F[a-1][b-1]*(a-1)*(b-1)
        return temp%mod
    
    F[A][B]=1
    for i in range(A+1,C+1):
        F[i][B]=(F[i-1][B]*B)%mod
    for j in range(B+1,D+1):
        F[A][j]=(F[A][j-1]*A)%mod
    

    
    for i in range(A+1,C+1):
        for j in range(B+1,D+1):
            F[i][j]=calc(i,j)
            
 
    
    print(F[C][D])



    
    
 

main()