Problem:
Score : 400 points

Problem Statement

We have a large square grid with H rows and W columns.
Iroha is now standing in the top-left cell.
She will repeat going right or down to the adjacent cell, until she reaches the bottom-right cell.

However, she cannot enter the cells in the intersection of the bottom A rows and the leftmost B columns. (That is, there are A×B forbidden cells.) There is no restriction on entering the other cells.

Find the number of ways she can travel to the bottom-right cell.

Since this number can be extremely large, print the number modulo 10^9+7.

Constraints

1 ≦ H, W ≦ 100,000

1 ≦ A < H

1 ≦ B < W

Input

The input is given from Standard Input in the following format:

H W A B

Output

Print the number of ways she can travel to the bottom-right cell, modulo 10^9+7.

Sample Input 1

2 3 1 1

Sample Output 1

2

We have a 2×3 grid, but entering the bottom-left cell is forbidden. The number of ways to travel is two: "Right, Right, Down" and "Right, Down, Right".

Sample Input 2

10 7 3 4

Sample Output 2

3570

There are 12 forbidden cells.

Sample Input 3

100000 100000 99999 99999

Sample Output 3

1

Sample Input 4

100000 100000 44444 55555

Sample Output 4

738162020

Solution in Kotlin (1.0.0) (judge verdict: Accepted)

fun main(args: Array<String>) = yorukatsu10f()

fun yorukatsu10f() {
    val (h, w, a, b) = readLine()!!.split(' ').map { it.toInt() }

    nCkInit()
    var answer = 0L

    for (i in 0 until w - b) {
        val nCk = nCk(b + i + (h - a) - 1, h - a - 1) * nCk(w - b + a - 2 - i, a - 1)
        answer += nCk
        answer %= MOD
    }

    println(answer)
}

private const val MOD = (1e9 + 7).toLong()

private const val MAX = 1000000
private val fac = LongArray(MAX)
private val fInv = LongArray(MAX)
private val inv = LongArray(MAX)

private fun nCk(n: Int, k: Int): Long {
    if (n < k) return 0
    if (n < 0 || k < 0) return 0
    return fac[n] * (fInv[k] * fInv[n - k] % MOD) % MOD
}

private fun nCkInit() {
    fac[0] = 1
    fac[1] = 1
    fInv[0] = 1
    fInv[1] = 1
    inv[1] = 1
    for (i in 2 until MAX) {
        fac[i] = fac[i - 1] * i % MOD
        inv[i] = MOD - inv[(MOD % i).toInt()] * (MOD / i) % MOD
        fInv[i] = fInv[i - 1] * inv[i] % MOD
    }
}