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 REP(i, a, b) for(int i = (a); i <= (b); i++)
#define PER(i, a, b) for(int i = (a); i >= (b); i--)
#define rep(i, a, b) for(int i = (a); i < (b); i++)
#define all(S) (S).begin(), (S).end()
#define pb push_back
#define mk make_pair
#define S second
#define F first
typedef long long ll;
typedef long double lf;
typedef pair<int, int> ii;
const int MOD = 1e9+7;
const int MAX = 2e5+5;

int expo(int b, int e = MOD-2) {
    int ans = 1;
    while(e) {
        if(e&1) ans = (1LL*ans*b)%MOD;
        b = (1LL*b*b)%MOD; e >>= 1;
    }
    return ans;
}

int main(int argc, char ** argv) {
	int N, M, a, b, ans = 0;
	scanf("%d%d", &N, &M);
	scanf("%d%d", &a, &b); a = N-a+1;
	vector<int> fact(MAX+1, 1);
	REP(i, 1, MAX) fact[i] = (1LL*fact[i-1]*i)%MOD;
	function<int(int,int)> calc = [&](int p, int n) {
	    return (1LL*fact[n]*expo((1LL*fact[n-p]*fact[p])%MOD))%MOD;
	};
	function<int(int,int)> comb = [&](int n, int m) {
	    return calc(n-1, n+m-2);
	};
	REP(i, b+1, M) {
	    int a1 = comb(a-1, i);
	    int a2 = comb(N-a+1, M-i+1);
	    ans = (ans+(1LL*a1*a2)%MOD)%MOD;
	}
	printf("%d\n", ans);
	return 0;
}