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

# A = 1, B = 1, C = 2, D = 2
# |0|0|0|
# |0|0|0|
# |0|0|0|
dp = [[0 for _ in range(D + 1)] for _ in range(C + 1)]  # (C + 1, D + 1)

# |0|0|0|
# |0|1|0|
# |0|0|0|
dp[A][B] = 1

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

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

# |0|1|2|    |0|1|4|    |0|1|3|
# |0|1|1| -> |0|1|1| -> |0|1|1|
# |0|0|0|    |0|0|0|    |0|0|0|
for i in range(A + 1, C + 1):
    for j in range(B + 1, D + 1):
        dp[i][j] = (i * dp[i][j - 1]) % MOD
        dp[i][j] = (dp[i][j] + (j * dp[i - 1][j]) % MOD) % MOD
        dp[i][j] = (dp[i][j] - ((i - 1) * (j - 1) * dp[i - 1][j - 1]) % MOD) % MOD

print(dp[C][D])