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 <iomanip>
#include <cstdio>
#include <string>
#include <cstring>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <complex>
#include <cmath>
#include <limits>
#include <cfloat>
#include <climits>
#include <ctime>
#include <cassert>
#include <numeric>
#include <functional>
using namespace std;

#define rep(i,a,n) for(int (i)=(a); (i)<(n); (i)++)
#define repq(i,a,n) for(int (i)=(a); (i)<=(n); (i)++)
#define repr(i,a,n) for(int (i)=(a); (i)>=(n); (i)--)
#define all(v) begin(v), end(v)
#define pb(a) push_back(a)
#define fr first
#define sc second
#define INF 2000000000
#define int long long int

#define X real()
#define Y imag()
#define EPS (1e-10)
#define EQ(a,b) (abs((a) - (b)) < EPS)
#define EQV(a,b) ( EQ((a).X, (b).X) && EQ((a).Y, (b).Y) )
#define LE(n, m) ((n) < (m) + EPS)
#define LEQ(n, m) ((n) <= (m) + EPS)
#define GE(n, m) ((n) + EPS > (m))
#define GEQ(n, m) ((n) + EPS >= (m))

typedef vector<int> VI;
typedef vector<VI> MAT;
typedef pair<int, int> pii;
typedef long long ll;

typedef complex<double> P;
typedef pair<P, P> L;
typedef pair<P, double> C;

int dx[]={1, -1, 0, 0};
int dy[]={0, 0, 1, -1};
int const MOD = 1000000007;

// ModInt begin

ll mod_pow(ll x, ll n) {return (!n)?1:(mod_pow((x*x)%MOD,n/2)*((n&1)?x:1))%MOD;}
struct ModInt {
    ll v;
    ModInt(ll a = 0) : v(((a%MOD) + MOD) % MOD) {}
    ModInt  operator+ ( const ModInt& b ) const {return (v + b.v) % MOD;}
    ModInt  operator- ( const ModInt& b ) const {return (v - b.v + MOD) % MOD;}
    ModInt  operator* ( const ModInt& b ) const {return (v * b.v) % MOD;}
    ModInt  operator/ ( const ModInt& b ) const {return (v * mod_pow(b.v, MOD-2)) % MOD;}
};

bool    operator==(ModInt  a, ModInt b) {return a.v == b.v;}
ModInt& operator+=(ModInt& a, ModInt b) {return a = a + b;}
ModInt& operator-=(ModInt& a, ModInt b) {return a = a - b;}
ModInt& operator*=(ModInt& a, ModInt b) {return a = a * b;}
ModInt& operator/=(ModInt& a, ModInt b) {return a = a / b;}
ostream& operator<<(ostream& out, ModInt  a) {return out << a.v;}
istream& operator>>(istream& in,  ModInt& a) {
    ll v; in >> v;
    a = ModInt(v);
    return in;
}

// ModInt end

ModInt fact[200010];

ll comb(int n, int r) {
    if(n < r) return 0;
    ModInt ret = fact[n];
    ret /= fact[n-r];
    ret /= fact[r];
    return ret.v;
}

signed main() {
    fact[0] = 1;
    rep(i,1,200010) fact[i] = fact[i-1] * i;

    int h, w, a, b; cin >> h >> w >> a >> b;
    ModInt ans = 0;
    repq(i,1,w-b) {
        ModInt va = comb(h-a+b+i-2, b+i-1);
        ModInt vb = comb(w-b-i+1+a-2, a-1);
        ans += va * vb;
    }
    cout << ans << endl;
    return 0;
}