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 <iostream>
#include <string>
#include <algorithm>
#include <cstdio>
#include <vector>
#include <queue>
#include <set>
#include <numeric>
#include <cmath>

using namespace std;

typedef long long int ll;

#define all(x) x.begin(),x.end()

const ll mod = 1e9+7;
const ll INF = 1e9;
const ll MAXN = 1e9;

vector<ll> fact,inv;

ll f(ll a,ll b, ll p)  // a^b mod pを求める
{
	if (b == 0)
	{
		return 1;
	}else if (b % 2 == 0)
	{
		ll d = f(a,b/2,p);  //a^b = a^(2k) = (a^k)^2  k = b / 2
		return d * d % mod;
	}else
	{
		return a * f(a,b-1,p) % mod;
	}
}

void init(ll n)
{
	fact.resize(n+1,1);
	inv.resize(n+1,1);

	for(int i = 2; i <= n; i++)  //　nまでの階乗を求める
	{
		fact[i] = fact[i-1] * i % mod;
	}
	for(int i = 2; i <= n; i++)  //a^(-1) mod p = a^(p-2) mod p  ここをもう少し早くできるぽい
	{
		inv[i] = f(fact[i],mod-2,mod) % mod;
	}
}

ll comb(ll n,ll k)  //n_C_k
{
	if (n < k)
	{
		return 0;
	}
	return (fact[n] * inv[k] % mod) * inv[n-k] % mod;
}


int main()
{
	ll h,w,a,b;
	cin>>h>>w>>a>>b;

	init(h+w+1);

	ll ans = 0;
	for(int x = 1; x <= h-a; x++){
		ans += comb(x-1+b-1,x-1) * comb(h-x+w-b-1,h-x);
		ans %= mod;
		//cout << ans << endl;
	}

	cout << ans << endl;

	return 0;
}