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 Kotlin (1.3.71) (judge verdict: Wrong Answer)



import java.util.*

fun main() {
    val sc = Scanner(System.`in`)
    val mod = 998244353.toLong()

    val a = sc.nextInt()
    val b = sc.nextInt()
    val c = sc.nextInt()
    val d = sc.nextInt()

    val dp = Array(c+1){LongArray(d+1){-1L} }

    fun solve(r:Int, c:Int): Long {
        if(r < a || c < b)
            return 0L
        if(r==a && c == b) {
            return 1L
        }
        if(dp[r][c] != -1L) {
            return dp[r][c]
        }
        // column 붙인 경우
        var cnt = 0L
        cnt += solve(r-1, c) * c
        cnt %= mod
        cnt += solve(r, c - 1) * r
        cnt %= mod
        cnt = cnt - (r-1)*(c-1)*solve(r -1, c - 1) + mod
        cnt %= mod
        dp[r][c] = cnt
        return cnt
    }
    print(solve(c, d))
    //println(".")


    //[a,b] -> [c,d]
    //



}