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:
MOD = 998244353
a, b, c, d = map(int, input().split())
dp = [[0]*(d+1) for _ in range(c+1)]
dp[a][b] = 1
for i in range(a, c+1):
  for j in range(b, d+1):
    if i == a and j == b:
      continue
    if i == a:
      dp[i][j] = dp[i][j-1]*i
    elif j == b:
      dp[i][j] = dp[i-1][j]*j
    else:
      dp[i][j] = dp[i-1][j]+dp[i][j-1]
      dp[i][j] += dp[i-1][j-1]*(i-1)*(j-1)
      dp[i][j] += (dp[i-1][j]-dp[i-1][j-1]*(i-1))*(j-1)
      dp[i][j] += (dp[i][j-1]-dp[i-1][j-1]*(j-1))*(i-1)
    dp[i][j] %= MOD
ans = dp[i][j]
print(ans)