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;
#include <iostream>
#include <vector>
#include <cstdlib>
#include <cmath>
#include <math.h>
#include <sstream>
#include <numeric>
#include <cctype>
#include <bitset>
#include <cassert>
#include<algorithm>
//---------------------------------------------------------------
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep1(i, n) for (int i = 1; i <= (int)(n); i++)
#define rep2(i, s, n) for (int i = (s); i < (int)(n); i++)
#define SIZE 100005
#define INF 1e18
#define all(x) x.begin(),x.end()
#define fi first
#define se second
#define vec vector
using pint = pair<int,int>;
using ll = long long;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
//---------------------------------------------------------------
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
const double pi = acos(-1);
//---------------------------------------------------------------

//↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓

const long long MAX = 1001000;
const long long MOD = 1000000007;

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

// 二項係数計算
long long com(long long n, long long k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

int main() {
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

    ll h,w,a,b;
    cin>>h>>w>>a>>b;

    COMinit();

    ll ans=0;
    for(int i=b+1;i<=w;i++){
        ll cur=com(h-1-a-1+i,h-a-1)*com(a-1+w-i,a-1);
        cur=cur%MOD;
        ans+=cur;
        ans=ans%MOD;
    }

    cout<<ans<<endl;
    return 0;
    
}