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 = [[-1 for i in range(d+1)] for j in range(c+1)]
for i in range(a+1):
    for j in range(b+1):
        dp[i][j] = 1
def doubling(n, m, modulo=0):
    y = 1
    base = n
    tmp = m
    while tmp:
        if tmp % 2 == 1:
            y *= base
            if modulo > 0:
                y %= modulo
        base *= base
        if modulo > 0:
            base %= modulo
        tmp //= 2
    return y
for i in range(b, d):
    dp[a][i+1] = dp[a][i] * a
    dp[a][i+1] %= mod
for i in range(a, c):
    dp[i+1][b] = dp[i][b] * b
    dp[i+1][b] %= mod
for i in range(a+1, c+1):
    for j in range(b+1, d+1):
        dp[i][j] = j*dp[i-1][j] + i*dp[i][j-1] - (i-1)*(j-1)*dp[i-1][j-1]
        dp[i][j] %= mod

print(dp[c][d])