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>
#include <string.h>
#include <stdbool.h>

#define S_MAX 100
#define N_MAX 100000
#define M_MAX 200000
#define INF 10000000
#define MOD 1000000007
#define SMAP(a, b) ((a)!=(b))&&((a)^=((b)^=((a)^= (b))))

typedef unsigned long long ull;
typedef   signed long long dll;

int n, m;
int h, w;
int i, j, k;
int a, b;

int arr[N_MAX * 2];

typedef struct {
	int in;
	int out;
} twoway;

void deb(){
	for (i = b; i < w; i++) {
		printf("[%d] %llu\n", i - b, arr[i - b]);
	}
	return;
}

ull bitpow(ull x, ull p){
	ull result = 1;
	while (p) {
		if (p & 1) {
			result *= x;
			result %= MOD;
		}
		p = (p >> 1);
		x *= x;
		x %= MOD;
	}
	return result;
}

ull dividual(ull b){
	return bitpow(b, MOD - 2);
}

ull frac(ull n){
	static ull dp[N_MAX * 2] = {1};
	if (dp[n]) return dp[n];
	return dp[n] = (frac(n - 1) * n) % MOD;
}

void solve(){
	ull result = 0;
	ull prev;
	
	for (i = b; i < w; i++) {
		prev = 1;
		prev *= frac(h - a - 1 + i);
		prev %= MOD;
		prev *= dividual(frac(h - a - 1));
		prev %= MOD;
		prev *= dividual(frac(i));
		prev %= MOD;
		prev *= frac(a - 1 + w - i - 1);
		prev %= MOD;
		prev *= dividual(frac(a - 1));
		prev %= MOD;
		prev *= dividual(frac(w - i - 1));
		prev %= MOD;

		arr[i - b] = prev;
	}

	for (i = b; i < w; i++) {
		result += arr[i - b];
		result %= MOD;
	}

	printf("%d\n", result);

	return;
}

int main (void) {
	scanf("%d%d%d%d", &h, &w, &a, &b);

	for (i = 0; i < N_MAX; i += 1000) frac(i);

	solve();
	return 0;
}