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 <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <vector>
#include <time.h>
#include <string>
#include <stack>
#include <set>
#include <map>
#include <iostream>
#include <algorithm>
using namespace std;
#define MP make_pair
#define PB push_back
typedef long long LL;
typedef pair<int, int> Pii;
const int inf = 0x3f3f3f3f;
const LL INF = (1uLL << 63) - 1;
const LL mod = 1000000007;
const int N = 2e5 + 5;
const double Pi = acos(-1.0);
const int maxn = 4e3 + 5;
LL fast_power(LL a,LL b){
    LL res = 1;
    while(b){
        if(b&1)res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}
LL fac[N];
LL inv[N];
void init(){
    fac[0] = 1;
    fac[1] = 1;
    for(int i = 2; i < N ; i++)fac[i] = fac[i-1] * i % mod;
    inv[N - 1] = fast_power(fac[N - 1], mod - 2);
    for(int i = N - 2; i >= 0; i--)inv[i] = inv[i + 1] * (i + 1) % mod;
}
LL cnt(int a,int b){
    a--;
    b--;
    a = a + b;
    return  fac[a] * inv[a - b] % mod * inv[b]  % mod;
}
int main() {
#ifdef local
    freopen("in", "r", stdin);
    //  freopen("w","w",stdout);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    init();
    int h,w,a,b;
    cin>>h>>w>>a>>b;
    LL res = 0;
    LL tmp = 0;
    LL last = 0;
    for(int i = w; i >= b + 1; i--){
        last = cnt(a + 1, w - i);
        tmp =  cnt(h - a,i) * (cnt(a + 1,w - i + 1) -  last + mod) % mod;
        res = (res + tmp) % mod;
    }
    cout<<res<<endl;
}