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 = lambda : sys.stdin.readline().rstrip()
sys.setrecursionlimit(max(1000, 10**9))
write = lambda x: sys.stdout.write(x+"\n")



a,b,c,d = map(int, input().split())

dp = [[0]*(d+1) for _ in range(c+1)]
dp[a][b] = 1
M = 998244353
for x in range(a+1, c+1):
    dp[x][b] = b*dp[x-1][b]
    dp[x][b] %= M
for y in range(b+1, d+1):
    dp[a][y] = a*dp[a][y-1]
    dp[a][y] %= M
for x in range(a+1, c+1):
    for y in range(b+1, d+1):
        dp[x][y] = dp[x-1][y]+dp[x][y-1]+(x-1)*dp[x][y-1]+(y-1)*dp[x-1][y]-(x-1)*(y-1)*dp[x-1][y-1]
        dp[x][y] %= M
print(dp[c][d])