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 <bits/stdc++.h>
using namespace std;

using VS = vector<string>;    using LL = long long;
using VI = vector<int>;       using VVI = vector<VI>;
using PII = pair<int, int>;   using PLL = pair<LL, LL>;
using VL = vector<LL>;        using VVL = vector<VL>;

#define ALL(a)  begin((a)),end((a))
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define SZ(a) int((a).size())
#define SORT(c) sort(ALL((c)))
#define RSORT(c) sort(RALL((c)))
#define UNIQ(c) (c).erase(unique(ALL((c))), end((c)))
#define FOR(i, s, e) for (int(i) = (s); (i) < (e); (i)++)
#define FORR(i, s, e) for (int(i) = (s); (i) > (e); (i)--)
#define debug(x) cerr << #x << ": " << x << endl
const int INF = 1e9;                          const LL LINF = 1e16;
const LL MOD = 1000000007;                    const double PI = acos(-1.0);
int DX[8] = { 0, 0, 1, -1, 1, 1, -1, -1 };    int DY[8] = { 1, -1, 0, 0, 1, -1, 1, -1 };

/* -----  2018/02/15  Problem: 058_arc_b / Link: https://arc058.contest.atcoder.jp/tasks/arc058_b?lang=en ----- */
/* ------問題------

縦 H マス、横 W マスのマス目があります。 いろはちゃんは、今一番左上のマス目にいます。 そして、右か下に1マス移動することを繰り返し、一番右下のマス目へと移動します。 ただし、下から A 個以内、かつ左から B 個以内のマス目へは移動することは出来ません。
移動する方法は何通りあるか求めてください。
なお、答えは非常に大きくなることがあるので、答えは 109+7 で割ったあまりを出力してください。

-----問題ここまで----- */
/* -----解説等-----



----解説ここまで---- */




const long long mod = 1e9 + 7;

long long modpow(long long a, long long b) {
	if (b == 0) return 1;
	return modpow(a * a % mod, b / 2) * (b & 1 ? a : 1) % mod;
}

long long modinv(long long a) {
	return modpow(a, mod - 2);
}

vector<long long> fact, inv_fact;

void init_fact(int n) {
	fact.resize(n);
	fact[0] = 1;
	for (int i = 1; i < n; i++) {
		fact[i] = i * fact[i - 1] % mod;
	}
	inv_fact.resize(n);
	inv_fact[n - 1] = modinv(fact[n - 1]);
	for (int i = n - 2; i >= 0; i--) {
		inv_fact[i] = (i + 1) * inv_fact[i + 1] % mod;
	}
}

long long nPr(int n, int r) {
	if (n < r || n < 0 || r < 0) return 0;
	return fact[n] * inv_fact[n - r] % mod;
}

long long nCr(int n, int r) {
	if (n < r || n < 0 || r < 0) return 0;
	return fact[n] * inv_fact[r] % mod * inv_fact[n - r] % mod;
}

long long nHr(int n, int r) {
	if (n == 0 && r == 0) return 1;
	return nCr(n + r - 1, r);
}
LL H, W, A, B;
LL ans = 0LL;
int main() {
	cin.tie(0);
	ios_base::sync_with_stdio(false);

	cin >> H >> W >> A >> B;
	init_fact(300000);
	FOR(h, 0, H - A) {
		ans += (nCr(B - 1 + h, B - 1) * nCr(W - 1 - B + H - 1 - h, W - 1 - B)) % MOD;
		ans %= MOD;
	}


	cout << ans << "\n";

	return 0;
}