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

#dp[i][j]たてにi回、横にj回追加したときの数
DP=[[0]*(D-B+1) for _ in range(C-A+1)]
DP[0][0]=1

for i in range(1,C-A+1):
    DP[i][0]=DP[i-1][0]*B
    DP[i][0]%=MOD
for i in range(1,D-B+1):
    DP[0][i]=DP[0][i-1]*A
    DP[0][i]%=MOD

for i in range(1,min(C-A,D-B)+1):
    for j in range(i,C-A+1):
        DP[j][i] = DP[j-1][i]*(B+i) + DP[j][i-1]*(A+j)-DP[j-1][i-1]*(B+i-1)*(A+j-1)
        DP[j][i]%=MOD
    for j in range(i+1,D-B+1):
        DP[i][j] = DP[i-1][j]*(B+j) + DP[i][j-1]*(A+i)-DP[i-1][j-1]*(B+j-1)*(A+i-1)
        DP[i][j]%=MOD


print(DP[-1][-1])