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 <bits/stdc++.h>
using namespace std;
typedef long long ll;

ll fact[200001];
ll finv[200001];
ll mod = 1e9+7;

ll mpow(ll a, ll b) {
    ll ret = 1;
    while (b > 0) {
        if (b&1) ret = ret * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ret;
}

ll c(ll n, ll k) {
    return fact[n] * finv[n-k] % mod * finv[k] % mod;    
}

int main() {
    int h, w, a, b; cin >> h >> w >> a >> b;
    fact[0] = fact[1] = 1;
    for (int i = 2; i <= h+w; i++) fact[i] = fact[i-1] * i % mod;
    finv[h+w] = mpow(fact[h+w], mod-2);
    for (int i = h+w-1; i >= 0; i--) finv[i] = finv[i+1] * (i+1) % mod;
    ll ans = 0;
    for (int i = b; i < w; i++) {
        ans = (ans + c(h-a-1+i, i) * c(a-1+w-i-1,w-i-1) % mod) % mod;
    }
    cout << ans << endl;
    return 0;
}