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: Wrong Answer)

/*
 * じょえチャンネル
 * 高評価・チャンネル登録よろしくおねがいします！
 * https://www.youtube.com/channel/UCRXsI3FL_kvaVL9zoolBfbQ
 */
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,n) for(int i=1;i<n;i++)
#define all(v) v.begin(),v.end()
#define inf (int)(3e18)
#define P pair<int,int>
#define mod (int)(1e9+7)
template<class T> inline void chmin(T &a, T b) {
	a = min(a, b);
}
template<class T> inline void chmax(T &a, T b) {
	a = max(a, b);
}
int mod_pow(int x, int y, int m = mod) {
	int res = 1;
	x %= m;
	while (y > 0) {
		if (y & 1) {
			res = res * x % m;
		}
		x = x * x % m;
		y >>= 1;
	}
	return res;
}
int perm[1000005];
void init_perm(){
	perm[0]=1;
	REP(i,1000005)perm[i]=perm[i-1]*i%mod;
}
int nCk(int x,int y){
	return perm[x]*mod_pow(perm[x-y],mod-2)%mod*mod_pow(perm[y],mod-2)%mod;
}

int H,W,A,B;
signed main(){
	cin>>H>>W>>A>>B;
	int ans=0;
	init_perm();
	for(int i=B+1;i<=W;i++){
		int p=nCk(i-1+H-A-2,i-1);
		if(i==B+1)p+=nCk(i-2+H-A-1,i-2);
		p%=mod;
		int q=nCk(A+W-i,A);
		ans+=p*q%mod;
		ans%=mod;
	}
	cout<<ans<<endl;
}