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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
 
 
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3") 
#pragma GCC optimization ("unroll-loops")
 
// #include <ext/pb_ds/detail/standard_policies.hpp>
using namespace std;
using namespace __gnu_pbds;
// const ll RANDOM = chrono::high_resolution_clock::now().time_since_epoch().count();
// gp_hash_table<int, int> table;
 
#define ll  long long
#define vii vector< ll >
#define pb push_back
#define pii pair<ll,ll>
#define ppi pair<pii,ll>
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define ins insert
#define ld long double
#define lp(l,r) for(i=l;i<=r;++i)
 
#define lpj(l,r) for(j=l;j<=r;++j)
#define lpr(l,r) for(i=r;i>=l;--i)
#define lprj(l,r) for(j=r;j>=l;--j)
 
#define swap(a,b){ auto _t=a;a=b,b=_t;}
#define flush() cout.flush()
#define popcount __builtin_popcount
#define die() return 0;
#define sz() size() 


 
 
#define TRACE
 
#ifdef TRACE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
    cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
    const char* comma = strchr(names + 1, ',');cerr.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
#else
#define trace(...)
#endif
 
#define MD          1000000007
#define M 			1000011
#define LN             14
#define fftmod      998244353
#define EPS         (double)1e-10
 
long long  md=MD;
ll inf=1e18;
 
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count()); 
 
 
inline ll exp(ll a,ll b){if(a==0&&b==0)return 1ll;if(a==0)return 0ll;ll r=1LL;while(b>0){if(b&1){r=(r%md)*(a%md);r=(r+md)%md;}b/=2;a=(a%md)*(a%md);a=(a+md)%md;}return (r+md)%md;}
ll gcd(ll a,ll b){if(b==0)return a;if(a==0)return b;return gcd(b,a%b);}
inline ll poww(ll a,ll b){ll r=1LL;while(b>0){if(b&1)r=r*a;b/=2;a=a*a;}return (ll)r;}
 
using cd=complex<double>;
const double PI=acos(-1);
 
inline ll rand(ll l,ll r){
    return l+rng()%(r-l+1);
 
}
ll fa[M],iv[M];
ll chose(ll n,ll k){
	if(n<k)return 0;
	ll r=fa[n];r=r*iv[k];r%=md;r=r*iv[n-k]%md;
	return r%md;
}

ll way(ll x1,ll y1,ll x2,ll y2){
	if(x1>x2||y1>y2)return 0;
	return chose(y2+x2-y1-x1,x2-x1);

}
int main(){
 
     
    ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
 
 
   // ifstream fin("input.txt");ofstream fout("outut.txt");
    ll t,i,j,m,n,s,z,k,p;string s1,s2;ll q;ll x,y;
    fa[0]=iv[0]=1;
    lp(1,M-1){
    	fa[i]=fa[i-1]*i%md;fa[i]%=md;
    	iv[i]=iv[i-1]*exp(i,md-2)%md;iv[i]%=md;
    }
    ll h,w,a,b;

    cin>>h>>w>>a>>b;
    p=0;
    for(i=h-a;i>=1;--i){
    	k=way(1,1,i,b);k%=md;
    	// trace(1,1,i,b);
    	k=k*way(i,b+1,h,w)%md;
    	// trace(i,w-b,h,w);
    	p+=k;
    	// trace(k);

    	p%=md;

    }
    cout<<p%md<<"\n";






    	






    

}