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

stdin = sys.stdin


def ni(): return int(ns())


def na(): return list(map(int, stdin.readline().split()))


def naa(N): return [na() for _ in range(N)]


def ns(): return stdin.readline().rstrip()  # ignore trailing spaces


A, B, C, D = na()

mod = 998244353

dp = [[0] * (D + 1) for _ in range(C + 1)]

dp[A][B] = 1
for a in range(A + 1, C + 1):
    dp[a][B] = (dp[a - 1][B] * B) % mod

for b in range(B + 1, D + 1):
    dp[A][b] = (dp[A][b - 1] * A) % mod

for a in range(A + 1, C + 1):
    for b in range(B + 1, D + 1):
        dp[a][b] = (dp[a - 1][b] * b + dp[a][b - 1] * a -
                    dp[a - 1][b - 1] * (a - 1) * (b - 1)) % mod

print(dp[C][D])