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: Time Limit Exceeded)

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

dp = Array.new(c + 1) {Array.new(d + 1) {[0, 0]}}

dp[a][b][0] = 1
(a+1).upto(c) {|i| dp[i][b][0] = dp[i-1][b][0] * b % M}
(b+1).upto(d) {|j| dp[a][j][0] = dp[a][j-1][0] * a % M}

(a+1).upto(c) do |i|
    (b+1).upto(d) do |j|
        dp[i][j][0] = (dp[i-1][j][0] + dp[i-1][j][1]) * j % M
        dp[i][j][1] = (dp[i][j-1][0] + dp[i][j-1][1] * i) % M
    end
end

p (dp[c][d][0] + dp[c][d][1]) % M