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 <iomanip>
#include <string>
#include <vector>
#include <queue>
#include <algorithm>
#include <utility>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <random>
#include <deque>

#define INF_LL 1e18
#define INF 1e9

#define REP(i, n) for(int i = 0;i < (n);i++)
#define FOR(i, a, b) for(int i = (a);i < (b);i++)
#define all(x) x.begin(),x.end()

using namespace std;

using ll = long long;
using PII = pair<int, int>;
using PLL = pair<ll, ll>;

const ll MOD = 1e9+7;

int dx[4] = {-1, 1, 0, 0};
int dy[4] = {0, 0, -1, 1};

template<typename T>
void chmax(T &a, T &b){
	a = max(a, b);
}

template<typename T>
void chmin(T &a, T &b){
	a = min(a, b);
}

class Union_find{
private:
	vector<int> par;
	vector<int> rank;
	int n;

public:
	Union_find(int a){
		n = a;
		for(int i = 0;i < n;i++){
			par.push_back(i);
			rank.push_back(0);
		}
	}

	int find(int x){
		if(par[x] == x){
			return x;
		}else{
			return par[x] = find(par[x]);
		}
	}

	void unite(int x, int y){
		x = find(x);
		y = find(y);
		if(x == y) return;

		if(rank[x] < rank[y]){
			par[x] = y;
		}else{
			par[y] = x;
			if(rank[x] == rank[y]) rank[x]++;
		}
	}

	bool same(int x, int y){
		return find(x) == find(y);
	}
};

class RMQ{
	int n;
	vector<int> dat;
public:
	RMQ(int n_){
		n = 1;
		while(n < n_) n *= 2;
		REP(i, n*2-1){
			dat.push_back(INF);
		}
	}

	void update(int x, int i){
		i += n-1;
		dat[i] = x;
		while(i > 0){
			i = (i-1)/2;
			dat[i] = min(dat[i*2+1], dat[i*2+2]);
		}
	}

	int query(int a, int b, int l, int r, int x){
		if(a <= l && r <= b) return dat[x];
		if(r <= a || b <= l) return INF;

		int res = INF;
		res = min(query(a, b, l, (l+r)/2, x*2+1), query(a, b, (l+r)/2, r, x*2+2));
		return res;
	}
	int query(int a, int b){
		return query(a, b, 0, n, 0);
	}
};

const ll mod = 1e9+7;
ll frac[214514] = {};

ll pow(ll a, ll b, ll m){
	if(b == 0) return 1;
	if(b == 1) return a%m;
	ll res = 1;
	if(b % 2) res = (res*a)%m;
	ll calc = pow(a, b/2, m);
	res = (res*calc)%m;
	res = (res*calc)%m;
	return res;
}

ll comb(ll a, ll b){
	ll res = frac[a]*pow(frac[b], mod-2, mod)%mod;
	res = (res*pow(frac[a-b], mod-2, mod))%mod;
	return res;
}

int main(void){
	ll H, W, A, B;
	cin >> H >> W >> A >> B; H--; W--;
	frac[0] = 1;
	FOR(i, 1, 214514){
		frac[i] = (frac[i-1] * i)%mod;
	}
	if(A == 0 || B == 0){
		cout << comb(H+W, H) << endl;
		return 0;
	}
	ll res = 0;
	for(int w = B;w <= W;w++){
		res = (res + comb(w+H-A, w)*comb(A+W-w-1, A-1))%mod;
	}
	cout << res << endl;
	return 0;
}