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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

#define MOD	((long long) 1e9 + 7)
#define N	200000

int ii[N + 1];

int mult(int a, int b) {
	return (int) ((long long) a * b % MOD);
}

int power(int a, int b) {
	int p;

	if (b == 0)
		return 1;
	p = power(a, b / 2);
	p = mult(p, p);
	if (b % 2 == 1)
		p = mult(p, a);
	return p;
}

int choose(int n, int k) {
	return k == 0 ? 1 : mult(mult(choose(n - 1, k - 1), n), ii[k]); 
}

int inv(int a) {
	return power(a, MOD - 2);
}

void init() {
	int a;

	for (a = 1; a <= N; a++)
		ii[a] = inv(a);
}

int main() {
	int h, w, a, b, i;
	int ch1, ch2, cnt;

	init();
	scanf("%d%d%d%d", &h, &w, &a, &b);
	h--, w--;
	ch1 = choose(b - 1, b - 1);
	ch2 = choose(w - b + h, w - b);
	for (i = 0; i <= h - a; i++) {
		cnt = (cnt + mult(ch1, ch2)) % MOD;
		ch1 = mult(mult(ch1, b - 1 + i + 1), ii[i + 1]);
		ch2 = mult(mult(ch2, ii[w - b + h - i]), h - i);
	}
	printf("%d\n", cnt);
	return 0;
}