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)

import java.math.BigInteger

val MOD = 1000000007L
val BMOD = BigInteger.valueOf(MOD)

fun main(args: Array<String>) {
    val (h, w, a, b) = readLine()!!.split(' ').map(String::toLong)
    var ans = 0L
    val n1 = h - a - 1 + b
    val n2 = w - b - 1 + a
    var k1 = b
    var k2 = a
    var c1 = c(n1, k1)
    var c2 = c(n2, k2)
    while (true) {
        ans = (ans + mult(c1, c2)) % MOD
        k1++; k2++
        if (k1 > n1 || k2 > n2) break
        c1 = div(mult(c1, n1 - k1 + 1), k1)
        c2 = div(mult(c2, n2 - k2 + 1), k2)
    }
    println(ans)
}

fun mult(a : Long, b : Long) = (a * b) % MOD
fun div(a : Long, b : Long) = mult(a, BigInteger.valueOf(b).modInverse(BMOD).toLong())

fun c(n : Long, k : Long) : Long {
    var ch = 1L
    var zn = 1L
    for (i in (n - k + 1)..n) ch = mult(ch, i)
    for (i in 1..k) zn = mult(zn, i)
    return div(ch, zn)
}