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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define task "atcoder"
#define rep(i,a,b) for(int i=(a),_b=(b);i<=_b;++i)
#define per(i,b,a) for(int i=(b),_a=(a);i>=_a;--i)
#define BIT(x, i) (((x) >> (i)) & 1)
#define lb lower_bound
#define ub upper_bound
#define lla(v) (v).begin(),(v).end()
#define zs(v) int((v).size())
#define fi first
#define se second
#define pb push_back
#define mp make_pair
typedef long double ld;
typedef long long ll;
typedef pair<int,int> pii;
const ll mod=1000000007; /// 998244353
const int base=311;
const int N=1e5+5;

ll fact[2*N];

ll pw(ll n, ll k)
{
    if(k==1) return n;
    if(k==0) return 1;
    ll t=pw(n,k/2);
    if(k&1) return t*t%mod*n%mod;
    else return t*t%mod;
}
ll C(ll n, ll k)
{
    ll ans=fact[n-k]*fact[k]%mod;
    ans=pw(ans,mod-2);
    return (ans*fact[n]%mod);
}
void solved()
{
    fact[0]=1;
    rep(i,1,200000) fact[i]=fact[i-1]*i%mod;
    int n,m,a,b;
    cin>>n>>m>>a>>b;
    ll ans=0;
    for(int i=b+1;i<=m;++i)
    {
        ll ways=C(n-a+i-2,n-a-1);
//        cout<<ways<<endl;
//        cout<<n-a+i<<' '<<n-a<<' '<<i<<endl;
        ll tam=C(a+m-i+1-2,a-1);
        ans=(ans+ways*tam%mod)%mod;
    }
    cout<<ans;
}
int main()
{
    ios_base::sync_with_stdio(0); cin.tie(0);
    if(fopen(task".inp","r"))
    {
        freopen(task".inp","r",stdin);
        freopen(task".out","w",stdout);
    }
    solved();
}