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<iostream>
using namespace std;
typedef long long ll;
const ll Mod = 1e9 + 7;
const int N = 200005;
ll Inv[2*N], Fac[2*N];
ll C(ll n, ll m) {
    if(!m || (n == m)) return 1;
    return Fac[n]*Inv[n-m]%Mod*Inv[m]%Mod;
}
ll Cal(ll x, ll y) {
    return C(x+y-2,y-1);
}
ll ex_gcd(ll a, ll b, ll &x, ll &y) {
    ll r, x1, y1;
    if(!b) {
        x = 1, y = 0; return a;
    }
    r = ex_gcd(b,a%b,x1,y1);
    x = y1,  y = x1-a/b*y1;
    return r;
}
ll inv(ll a, ll n) {
    ll x, y;
    if(ex_gcd(a,n,x,y) == 1) return (x+n)%n;
    else return -1;
}
int main() {
    ll H, W, A, B, ans = 0, n = N+N/2;
    cin >> H >> W >> A >> B;
    Fac[1] = 1;
    for(int i = 2; i <= n; ++i) {
        Fac[i] = (Fac[i-1]*i)%Mod;
    }
    for(int i = 1; i <= n; ++i) Inv[i] = inv(Fac[i],Mod);
    
    for(ll i = B+1; i <= W; ++i) {
        ans = (ans + Cal(H-A,i)*Cal(A,W-i+1)) % Mod;
    }
    cout << ans << '\n';
return 0;
}