Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

1 1 2 2

Sample Output 1

3

Any two of the three squares other than the bottom-left square can be painted black.

Sample Input 2

2 1 3 4

Sample Output 2

65

Sample Input 3

31 41 59 265

Sample Output 3

387222020

Solution in Ruby (2.7.1) (judge verdict: Accepted)

END { load __FILE__ unless $stdin.eof? }

a, b, c, d = gets.split.map!(&:to_i)
M = 998244353

$dp = Array.new(c + 1) { Array.new(d + 1) }
a.upto(c) do |x|
  $dp[x][b] = b.pow(x - a, M)
end
b.upto(d) do |y|
  $dp[a][y] = a.pow(y - b, M)
end

def f(x, y)
  $dp[x][y] ||= (
    f(x - 1, y) * y + f(x, y - 1) * x - f(x - 1, y - 1) * (x - 1) * (y - 1)
  ) % M
end

puts f(c, d)