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: Runtime Error)

#include <bits/stdc++.h>
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)   FOR(i,0,n)
#define ALL(obj)   (obj).begin(),(obj).end()

#define debug(var)  do{std::cout << __LINE__ << "> " << #var << ": ";view(var);}while(0)
template<typename T> void view(T e){std::cout << e << std::endl;}
template<typename T> void view(const std::vector<T>& v){for(const auto& e : v){ std::cout << e << " "; } std::cout << std::endl;}
template<typename T> void view(const std::vector<std::vector<T> >& vv){ for(const auto& v : vv){ view(v); } }

using namespace std;
using ll = long long;
using vi = vector<int>;
using vl = vector<ll>;
using vvi = vector<vector<int>>;
using vvl = vector<vector<ll>>;

// modint: mod 計算を int を扱うように扱える構造体
template<int MOD> struct Fp {
    long long val;
    constexpr Fp(long long v = 0) noexcept : val(v % MOD) {
        if (val < 0) v += MOD;
    }
    constexpr int getmod() { return MOD; }
    constexpr Fp operator - () const noexcept {
        return val ? MOD - val : 0;
    }
    constexpr Fp operator + (const Fp& r) const noexcept { return Fp(*this) += r; }
    constexpr Fp operator - (const Fp& r) const noexcept { return Fp(*this) -= r; }
    constexpr Fp operator * (const Fp& r) const noexcept { return Fp(*this) *= r; }
    constexpr Fp operator / (const Fp& r) const noexcept { return Fp(*this) /= r; }
    constexpr Fp& operator += (const Fp& r) noexcept {
        val += r.val;
        if (val >= MOD) val -= MOD;
        return *this;
    }
    constexpr Fp& operator -= (const Fp& r) noexcept {
        val -= r.val;
        if (val < 0) val += MOD;
        return *this;
    }
    constexpr Fp& operator *= (const Fp& r) noexcept {
        val = val * r.val % MOD;
        return *this;
    }
    constexpr Fp& operator /= (const Fp& r) noexcept {
        long long a = r.val, b = MOD, u = 1, v = 0;
        while (b) {
            long long t = a / b;
            a -= t * b; swap(a, b);
            u -= t * v; swap(u, v);
        }
        val = val * u % MOD;
        if (val < 0) val += MOD;
        return *this;
    }
    constexpr bool operator == (const Fp& r) const noexcept {
        return this->val == r.val;
    }
    constexpr bool operator != (const Fp& r) const noexcept {
        return this->val != r.val;
    }
    friend constexpr ostream& operator << (ostream &os, const Fp<MOD>& x) noexcept {
        return os << x.val;
    }
    friend constexpr istream& operator >> (istream &is, Fp<MOD>& x) noexcept {
        return is >> x.val;
    }
    friend constexpr Fp<MOD> modpow(const Fp<MOD> &a, long long n) noexcept {
        if (n == 0) return 1;
        auto t = modpow(a, n / 2);
        t = t * t;
        if (n & 1) t = t * a;
        return t;
    }
};

// 二項係数ライブラリ
template<class T> struct BiCoef {
    vector<T> fact_, inv_, finv_;
    constexpr BiCoef() {}
    constexpr BiCoef(int n) noexcept : fact_(n, 1), inv_(n, 1), finv_(n, 1) {
        init(n);
    }
    constexpr void init(int n) noexcept {
        fact_.assign(n, 1), inv_.assign(n, 1), finv_.assign(n, 1);
        int MOD = fact_[0].getmod();
        for(int i = 2; i < n; i++){
            fact_[i] = fact_[i-1] * i;
            inv_[i] = -inv_[MOD%i] * (MOD/i);
            finv_[i] = finv_[i-1] * inv_[i];
        }
    }
    constexpr T com(int n, int k) const noexcept {
        if (n < k || n < 0 || k < 0) return 0;
        return fact_[n] * finv_[k] * finv_[n-k];
    }
    constexpr T fact(int n) const noexcept {
        if (n < 0) return 0;
        return fact_[n];
    }
    constexpr T inv(int n) const noexcept {
        if (n < 0) return 0;
        return inv_[n];
    }
    constexpr T finv(int n) const noexcept {
        if (n < 0) return 0;
        return finv_[n];
    }
};

const int MOD = 1000000007;
using mint = Fp<MOD>;
BiCoef<mint> bc;

void Main() {
	int h, w, a, b; cin >> h >> w >> a >> b;
	bc.init(110000);
	mint res = 0;
    mint last_com2 = 0;
    for (int i = w - b - 1; i >= 0; --i) {
        int y = h - a;
        int x = b + i + 1;
        int y2 = a + 1;
        int x2 = w - b - i;
        mint com = bc.com(y + x - 1 - 1, x - 1);
        mint com2 = bc.com(y2 + x2 - 1 - 1, x2 - 1);
        // debug(com);
        // debug(com2);
        res += com * (com2 - last_com2);
        last_com2 = com2;
    }
    cout << res << endl;
}

int main() {
    cin.tie(nullptr);
	ios_base::sync_with_stdio(false);
	cout << fixed << setprecision(15);
	Main();
}