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 <vector>

using namespace std;
using ll = long long;
using cll = const long long;
using vec = vector<ll>;

ll pow(ll b, ll p, ll mod){
    ll ret = 1LL;
    ll cum = b;
    while (p){
        if (p & 1) ret = ret * cum % mod;
        cum = cum * cum % mod;
        p >>= 1;
    }
    return ret;
};

class ModuloMath{
    static constexpr ll mod = 1e9+7;
public:
    ModuloMath(ll n) : maxN(n){
        inv.resize(maxN + 1, 1);
        fac.resize(maxN + 1, 1);
        invfac.resize(maxN + 1, 1);
        for (ll i = 2; i <= maxN; ++i) inv[i] = mod - (mod / i) * inv[mod % i] % mod;
        for (ll i = 2; i <= maxN; ++i) fac[i] = fac[i - 1] * i % mod;
        invfac[maxN] = pow(fac[maxN], mod - 2, mod);
        for (ll i = maxN - 1; i != 0; --i) invfac[i] = invfac[i + 1] * (i + 1LL) % mod;
    }
    ll nCr(ll n, ll r){return fac[n] * invfac[r] % mod * invfac[n - r] % mod;}
    ll nPr(ll n, ll r){return fac[n] * invfac[n - r] % mod;}
    ll nHr(ll n, ll r){return nCr(n + r - 1, r);}
    // nHr 重複組み合わせ。n種類のものから重複を許してr個選ぶ場合の数。n種類の玉をr個（n < rで良い）選ぶときのパターンの数など。

private:
    ll maxN;
    vec inv;
    vec fac;
    vec invfac;
};

int main() {
    ll mod = 1e9+7;
    ll H, W, A, B;
    cin >> H >> W >> A >> B;
    ModuloMath mm(H + W + 1);
    ll total = mm.nCr(H + W - 2, H - 1);
    long long w = W - B;
    for (long long h = A; h != 0; --h){
        ll tmp = mm.nCr(h + w - 2, h - 1) * mm.nCr(H - h + 1 + B - 2, H - h + 1 - 1) % mod;
        total = (mod + total - tmp) % mod;
    }
    cout << total << endl;
    return 0;
}