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 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn = 1 + 1e5;
const ll mod = 1e9 + 7;

typedef pair<int,int> II;

int h, w, a, b, dp[maxn];
ll gt[2 * maxn + 100], igt[2 * maxn + 100];
ll ans;

ll C(int n, int k) {
	return ((gt[n] * igt[k]) % mod * igt[n - k]) % mod;
}

ll Pow (int i, int j) {
	if (j == 0) return 1;
	if (j % 2) {
		return (i * Pow (i, j - 1)) % mod;
	}
	ll val = Pow (i, j / 2);
  return (val * val) % mod;
}

void calc () {
	gt[0] = 1;
	for (int i = 1; i <= 2 * maxn; ++i) {
		gt[i] = (gt[i - 1] * i) % mod;
	}
	igt[2 * maxn] = Pow (gt[2 * maxn], mod - 2);
  for (int i = 2 * maxn - 1; i >= 0; --i) {
    igt[i] = (igt[i + 1] * (i + 1) + mod) % mod;
  }
}

int main()
{
	#ifndef ONLINE_JUDGE
	freopen("abc.inp","r",stdin);freopen("abc.out","w",stdout);
//	freopen("abc.inp","r",stdin);freopen("abc.out","w",stdout);
	#endif //ONLINE_JUDGE
	ios_base::sync_with_stdio(0);
	cin.tie();
	cout.tie();
	calc();

	cin >> h >> w >> a >> b;
	int _a = h - a, _b = w - b;
	for (int i = 0; i < _b; ++i)
		dp[i] = C(_a + b + i - 1, _a - 1);
	for (int i = 0; i < _b; ++i) {
		ans += (C (a + i - 1, i) * dp[_b - i - 1] ) % mod;
		ans %= mod;
//		cout << C (a + i - 1, i) << " " << dp[_b - i - 1] << '\n';
	}
	cout << ans;
   return 0;
}