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 <string>
#include <algorithm>
using namespace std;

using ll = long long;

constexpr ll mod = 1e9+7;

ll C[200001];
ll C2[200001];

ll pow(ll x, ll y, ll m) {
    ll s = 1;
    while(y != 0) {
        if(y & 1) {
            s = s*x % m;
        }
        x = x*x % m;
        y >>= 1;
    }
    return s;
}

void init(int h, int w) {
    C[0] = 1;
    C2[0] = 1;
    for(int i=1; i<=h+w; ++i) {
        C[i] = C[i-1] * i % mod;
        C2[i] = pow(C[i], mod-2, mod);
    }
}

int main() {
    int h, w, a, b;
    cin >> h >> w >> a >> b;
    init(h, w);
    ll res = 0;
    for(int i=b; i<w; ++i) {
        ll tmp = ((C[h-a-1+i] * C2[h-a-1] % mod) * C2[i]) % mod;
        ll tmp2 = ((C[w+a-i-2] * C2[a-1]) % mod) * C2[w-1-i] % mod;
        res = (res + (tmp * tmp2 % mod)) % mod;
    }
    cout << res << endl;
}