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>

typedef long long ll;

ll mod=1000000007ll;
ll f[2010101], rf[2010101];

ll inv(ll x) {
    ll res = 1;
    ll k = mod - 2;
    ll y = x;
    while (k) {
        if (k & 1) res = (res * y) % mod;
        y = (y * y) % mod;
        k /= 2;
    }
    return res;
}
void init() {
	int i;
    f[0] = 1;
	for(i=1;i<2010101;i++){f[i] = (f[i - 1] * i) % mod;}
    rf[2010101 - 1] = inv(f[2010101 - 1]);
	for(i=2010101-2;i>=0;i--){
    rf[i] = (rf[i + 1] * (i + 1)) % mod;
	}
}
//---------------------------------------------------------------------------------------------------
ll C(int n, int k) {
    if (k < 0 || n < k) return 0;

    ll a = f[n]; // = n!
    ll b = rf[n - k]; // = (n-k)!
    ll c = rf[k]; // = k!

    ll bc = (b * c) % mod;

    return (a * bc) % mod;
}

int main(){

	init();

	int H,W,A,B;

	scanf("%d %d %d %d",&H,&W,&A,&B);

	int i;

	ll ans=0ll;

	for(i=0;i<H-A;i++){
	ll add=C(B-1+i,B-1)*C(H-1-B+W-1-i,W-1-B);
	add%=mod;
	ans+=add;
	ans%=mod;
	}
	printf("%lld\n",ans);

	return 0;
}