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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N = 2e5;
const int mod = 1e9+7;
long long g[N];

long long binpow(int a,int x)
{
	long long res;
	if(x == 0) return 1;
	res = binpow(a,x/2) % mod;
	if(x % 2 == 0) return res*res % mod;
	else return ((a*res) % mod * res) % mod;
}
 
long long tohop(int k,int n)
{
	if(k < 0 || n < 0 || k > n) return 0;
	long long res = (g[n-k]*g[k]) % mod;
	return (g[n] * binpow(res,mod-2) ) % mod;
}

long long path(int x,int y,int x1,int y1)
{
	x1 = x1-x+1;
	y1 = y1-y+1;

	return tohop(x1-1,x1+y1-2);
}

signed main()
{
	int H,W,A,B;
	cin>>H>>W>>A>>B;
	g[0] = 1;
	

	for(int i=1;i<N;i++)
	{
		g[i] = ( g[i-1]*i ) % mod;
	}
	// cin>>H>>W>>A>>B;
	int ans = 0;
	
	for(int i=1;i<=H-A;i++)
	{
		// cout<<i<<' '<<B<<'\n';
		ans = ( ans + ( path(1,1,i,B)*path(i,B+1,H,W) ) % mod ) % mod;
	}

	cout<<ans;


}