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 endl '\n'

using namespace std;
template<class T, class T2> inline void chkmax(T &x, const T2 &y) { if(x < y) x = y; }
template<class T, class T2> inline void chkmin(T &x, const T2 &y) { if(x > y) x = y; }
const int MAXN = (1 << 20);
const int mod = (int)1e9 + 7;

int pw(int x, int b)
{
	int ret = 1;
	while(b)
	{
		if(b & 1) ret = (ret * 1ll * x) % mod;
		x = (x * 1ll * x) % mod;
		b >>= 1;
	}

	return ret;
}

int h, w, a, b;

void read()
{
	cin >> h >> w >> a >> b;
}

int fact[MAXN], inv_fact[MAXN];
inline int C(int n, int k) { if(n < 0 || k < 0 || n < k) return 0; return (((fact[n] * 1ll * inv_fact[k]) % mod) * 1ll * inv_fact[n - k]) % mod; }
inline int f(int x, int y) { return C(x + y - 2, y - 1); }

/*

   1 1 1 1 1 1 1 1
   1 2 3 4 5 
   1 3 6 10
   1 4
   1 5
   1
   1
   1

*/

void solve()
{
	fact[0] = 1; for(int i = 1; i < MAXN; i++) fact[i] = (fact[i - 1] * 1ll * i) % mod;
	inv_fact[MAXN - 1] = pw(fact[MAXN - 1], mod - 2); for(int i = MAXN - 2; i >= 0; i--) inv_fact[i] = (inv_fact[i + 1] * 1ll * (i + 1)) % mod;	

	int answer = f(h, w);
	for(int i = 1; i <= b; i++)
		answer = (answer - (f(h - a, i) * 1ll * f(a, w - i + 1)) % mod + mod) % mod;
	
	cout << answer << endl;
}

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);

	read();
	solve();
	return 0;
}