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>
using namespace std;
#define ll long long
#define fi first
#define se second
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define endl '\n'
#define ld long double
typedef pair<ll,ll> pii;
const int maxn=200005;
const int mod=1e9+7;
ll qp(ll a,ll b)
{
    ll ret=1;
    while(b)
    {
        if(b&1) ret=ret*a%mod;
        b>>=1;
        a=a*a%mod;
    }
    return ret;
}
ll jc[maxn],inv[maxn];
void init()
{
    jc[0]=1;
    for(int i=1;i<=200004;i++) jc[i]=jc[i-1]*i%mod;
    inv[200004]=qp(jc[200004],mod-2);
    for(int i=200003;i>=0;i--) inv[i]=inv[i+1]*(i+1)%mod;
}
ll C(int x,int y)
{
    return jc[x]*inv[x-y]%mod*inv[y]%mod;
}
ll calc(int x,int y)
{
    return C(x+y-2,x-1);
}
int main()
{
    #ifdef local
    freopen("in.txt","r",stdin);
    #endif // local
    IOS;
    init();
    int h,w,a,b;
    cin>>h>>w>>a>>b;
    ll ans=0;
    for(int i=h-a+1;i<=h;i++)
    {
        ans+=calc(i,b)*calc(h-i+1,w-b)%mod;
        ans%=mod;
    }
    ans=calc(h,w)-ans;
    ans=(ans+mod)%mod;
    cout<<ans<<endl;
}