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
dpx=[[0 for i in range(D+1)] for j in range(C+1)]
dpy=[[0 for i in range(D+1)] for j in range(C+1)]
dpx[C][D]=1
dpy[C][D]=1
for i in range(D-1,B-1,-1):
    dpx[C][i]=(dpy[C][i+1]*C)%mod
    dpy[C][i]=(dpy[C][i+1]*C)%mod
for j in range(C-1,A-1,-1):
    dpx[j][D]=(dpx[j+1][D]*D)%mod
    dpy[j][D]=(dpx[j+1][D])%mod

for i in range(C-1,A-1,-1):
    for j in range(D-1,B-1,-1):
        dpx[i][j]=((dpx[i+1][j]*j)%mod+(dpy[i][j+1]*i)%mod)%mod
        dpy[i][j]=(dpx[i+1][j]+(dpy[i][j+1]*i)%mod)%mod

print(dpx[A][B])