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>
#define FOR(i, begin, end) for(int i=(begin);i<(end);i++)
#define REP(i, n) FOR(i,0,n)
#define IFOR(i, begin, end) for(int i=(end)-1;i>=(begin);i--)
#define IREP(i, n) IFOR(i,0,n)
#define SORT(a) sort(a.begin(), a.end())
#define REVERSE(a) reverse(a.begin(), a.end())
#define int long long
#define INF 1000000000000000
using namespace std;

typedef vector<int> vec;
typedef vector<vec> mat;
typedef pair<int, int> Pii;

template<typename T>
void readvec(vector<T> &a);
void readindex(vector<int> &a);

int fastpow(int x, int n, int m){
    int a = 1;
    IREP(i, 64){
        a = (a * a) % m;
        if(((n >> i) & 1) == 1) a = (a * x) % m;
    }
    return a;
}

class Combination
{
public:

    vec fact, invfact;
    int MAX_N, mod;

    Combination(int MAX_N, int mod): MAX_N(MAX_N), mod(mod) {
        fact = vec(MAX_N + 1);
        invfact = vec(MAX_N + 1);
        fact[0] = 1;
        FOR(i, 1, MAX_N + 1){
            fact[i] = (fact[i - 1] * i) % mod;
        }
        invfact[MAX_N] = fastpow(fact[MAX_N], mod - 2, mod);
        IREP(i, MAX_N){
            invfact[i] = (invfact[i + 1] * (i + 1)) % mod;
        }
    }

    int comb(int n, int r){
        if(r > n || r < 0 || n < 0) return 0;
        int a = fact[n];
        a = (a * invfact[r]) % mod;
        a = (a * invfact[n - r]) % mod;
        return a;
    }
};

signed main(){

    int H, W, A, B; cin >> H >> W >> A >> B;
    int mod = 1000000007;

    Combination C(H + W, mod);
    
    int ans = C.comb(H + W - 2, W - 1);

    REP(i, B){
        int tmp = C.comb(i + H - A - 1, i) * C.comb(W + A - i - 2, W - i - 1);
        tmp %= mod;
        ans = (ans + (mod - tmp)) % mod;
    }

    cout << ans;
    
    return 0;
}


template<typename T>
void readvec(vector<T> &a){
    REP(i, a.size()){
        cin >> a[i];
    }
}
void readindex(vector<int> &a){
    REP(i, a.size()){
        cin >> a[i];
        a[i]--;
    }
}