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 Julia (1.4.0) (judge verdict: Time Limit Exceeded)

using DataStructures
const double = Float64
const int = Int64

parseint(x) = parse(int, x)
readint() = map(parseint, split(readline()))

MOD = 998244353
function main()
  A,B,C,D = readint()

  s0 = Array{int,2}(undef,C,D)
  s1 = Array{int,2}(undef,C,D)

  s0[A,B] = 1
  s1[A,B] = 0

  
  for i in A+1:C
    s0[i,B] = (s0[i-1,B]+s1[i-1,B])* B%MOD
    s1[i,B] = 0
  end

  for i in B+1:D
    s0[A,i] = 0
    s1[A,i] = (s0[A,i-1] + s1[A,i-1])*A%MOD
  end


  for j in B+1:D,i in A+1:C
    s0[i,j] = (s0[i-1,j]+s1[i-1,j])*j%MOD
    s1[i,j] = (s0[i,j-1]+s1[i,j-1]*i)%MOD
    
  end

  println((s0[C,D]+s1[C,D])%MOD)
  
end

main()