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 numpy as np
import math
import collections
import bisect


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

dp0 = [[1 for i in range(d + 1)] for j in range(c + 1)]
dp1 = [[0 for i in range(d + 1)] for j in range(c + 1)]
for i in range(c + 1):
    for j in range(d + 1):
        if j < b or i < a:
            dp0[i][j] = 0
            dp1[i][j] = 0
for i in range(a, c + 1):
    for j in range(b, d + 1):
        # print(i, j, dp[i][j])
        if a == i and b == j:
            continue
        dp0[i][j] = ((i * dp0[i][j - 1]) % 998244353 + dp1[i][j - 1]) % 998244353
        dp1[i][j] = (j * (dp0[i - 1][j] + dp1[i - 1][j]) % 998244353) % 998244353

print((dp0[c][d] + dp1[c][d]) % 998244353)