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 <stdio.h>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>
#include <map>
#include <set>

using namespace std;

typedef long long LL;
typedef pair<int , int> P2;
typedef pair<pair<int , int> , int> P3;
typedef pair<pair<int , int> , pair<int , int> > P4;
#define Fst first
#define Snd second
#define MP(a , b) make_pair((a) , (b))
#define M3P(a , b , c) make_pair(make_pair((a) , (b)) , (c))
#define M4P(a , b , c , d) make_pair(make_pair((a) , (b)) , make_pair((c) , (d)));
#define Mx(a , b) ((a) > (b) ? (a) : (b))
#define Mn(a , b) ((a) < (b) ? (a) : (b))

const double EPS = 1e-10;
const int INF = 2147483647;
const LL mod = 1e9 + 7;

int H , W , A , B;
LL con[200001];
LL rev[200001];
LL z[30];
LL ans;

LL f(LL a , LL b){
	return (rev[a] * rev[b] % mod) * con[a + b] % mod;
}

int main(){
	scanf("%d%d%d%d" , &H , &W , &A , &B);
	con[0] = 1;
	for(int i = 1 ; i < 200001 ; ++i){
		con[i] = (LL)i * con[i - 1];
		con[i] %= mod;
	}
	z[0] = con[200000];
	for(int i = 0 ; i < 29 ; ++i) z[i + 1] = z[i] * z[i] % mod;
	rev[200000] = 1;
	for(int i = 0 , j = mod - 2 ; j > 0 ; ++i , j /= 2){
		rev[200000] *= j % 2 == 1 ? z[i] : (LL)1;
		rev[200000] %= mod;
	}
	for(int i = 200000 ; i > 0 ; --i){
		rev[i - 1] = rev[i] * i % mod;
	}
	for(int i = 0 ; i < H - A ; ++i){
		ans += f(i , B - 1) * f(W - B - 1 , H - i - 1) % mod;
		ans %= mod;
	}
	printf("%lld\n" , ans);
	return 0;
}