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())

#ここまでのパテーン
#dp[y][x]=dp[y][x-1]*y +dp[y-1][x]*x

dp=[[0]*(d) for _ in range(c)]
dp[a-1][b-1]=1
mod=998244353

if a==c and b==d:
    print(1)
    exit()

if a==c:
    print((a**(d-b))%mod)
    exit()
if b==d:
    print((b**(c-a))%mod)
    exit()


for y in range(a-1,c):
    for x in range(b-1,d):
        
        dp[y][x]+=dp[y][x-1]*(y+1)+dp[y-1][x]*(x+1)-dp[y-1][x-1]*x*y
        dp[y][x]=dp[y][x]%mod
        #print(y,x,dp[y][x])

print(dp[-1][-1]%mod)