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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>
 
#define ll long long
#define ll128 __uint128_t
#define ld long double
#define pll pair <ll, ll>
 
#define vll vector <ll>
#define vld vector<ld>
#define vpll vector<pll>
 
#define vvll vector <vll>
 
#define rep(i, a, b) for(ll i = (ll)a; i < (ll)b; i++)
#define per(i, a, b) for(ll i = (ll)a - 1; i >= (ll)b; --i)
 
#define endl "\n"
#define pb push_back
#define pf push_front
 
#define all(v) (v).begin(), (v).end()
#define rall(v) (v).rbegin(), (v).rend()
 
#define sorta(v) sort(all(v))
#define sortd(v) sort(rall(v))
 
#define debug if (1)
#define log(val) debug {cout << "\n" << #val << ": " << val << "\n";}
 
#define ios ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define file(name) freopen(name".in", "r", stdin); freopen(name".out", "w", stdout);
#define FILE freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
 
#define mod (ll)(1e9 + 7)
#define inf (mod * mod)
 
using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;
 
ostream & operator << (ostream & out, vll & a) {
    for(auto i : a) out << i << " ";
    return out;
}
 
istream & operator >> (istream & in, vll & a) {
    for(auto &i : a) in >> i;
    return in;
}
 
const ll N = 2e5 + 5;
ll md = 1e9 + 7;
 
ll fact[N], revfact[N];
 
ll cnk1(ll n, ll k) {
    if (n < k || k == 0) return 1;
    return (((1LL * fact[n] % md * revfact[n - k] % md) % md) * revfact[k] % md) % md;
}
 
ll cnk(ll n, ll k) {
    return cnk1(n + k - 2, n - 1);
}
 
ll binpow (ll a, ll n, ll m) {
    ll res = 1;
    while (n > 0) {
        if (n & 1)
            res = res * a % m;
        a = a * a % m;
        n >>= 1;
    }
    return res;
}
 
void precalc() {
    fact[0] = 1;
    fact[1] = 1;
    rep(i, 2, N) {
        fact[i] = (1LL * fact[i - 1] % md * i % md) % md;
    }
    revfact[0] = 1;
    revfact[N - 1] = binpow(fact[N - 1], md - 2, md);
    for(ll i = N - 2; i >= 1; i--) {
        revfact[i] = (1LL * revfact[i + 1] % md * (i + 1) % md) % md;
    }
}
 
int main() {
    ll h, w, a, b;
    cin >> h >> w >> a >> b;
    precalc();
 
 
    ll curi = h - a;
    ll ans = 0;
    rep(j, b + 1, w + 1) {
        ans += (cnk(curi, j) * (cnk(a, w - j + 1)));
        ans %= md;
    }
    cout << ans << endl;
    return 0;
}