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:
'''
研究室PCでの解答
'''
import math
#import numpy as np
import queue
import bisect
from collections import deque,defaultdict
import heapq as hpq
from sys import stdin,setrecursionlimit
#from scipy.sparse.csgraph import dijkstra
#from scipy.sparse import csr_matrix
ipt = stdin.readline
setrecursionlimit(10**7)
mod = 998244353
dir = [(-1,0),(1,0),(0,-1),(0,1)]
alp = "abcdefghijklmnopqrstuvwxyz"


def main(a,b,c,d):
    dp = [[0]*(d+1) for i in range(c+1)]
    dp[a][b] = 1

    for i in range(a-1,c):
        for j in range(b-1,d):
            if i == a-1 and j == b-1:
                continue
            dpi = dp[i+1][j]*(i+1)+dp[i][j+1]*(j+1)-dp[i][j]*(i*j)
            dp[i+1][j+1] = dpi%mod

    return(dp[c][d])


    return None

if __name__ == '__main__':
    a,b,c,d = map(int,ipt().split())
    print(main(a,b,c,d))