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# (Mono 4.6.2.0) (judge verdict: Runtime Error)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.IO;

using Pair = System.Collections.Generic.KeyValuePair<int, int>;

class Program
{
	public Program() { }

	static void Main(string[] args)
	{
		new Program().prog();
	}
	Scanner cin;
	const int MOD = 1000000007;
	const int INF = int.MaxValue;
	const double EPS = 1e-7;
	const double PI = 3.1415926536;

	void prog()
	{
		cin = new Scanner();
		int[,] dir8 = new int[8, 2] { { -1, -1 }, { -1, 0 }, { -1, 1 }, { 0, -1 }, { 0, 1 }, { 1, -1 }, { 1, 0 }, { 1, 1 } };
		int[,] dir4 = new int[4, 2] { { -1, 0 }, { 0, -1 }, { 0, 1 }, { 1, 0 } };

		int H = cin.nextInt();
		int W = cin.nextInt();
		int A = cin.nextInt();
		int B = cin.nextInt();
		long ans = 0;
		long[] inv = new long[W - 2 + H - A + W - B + 1];
		long[] fac = new long[W - 2 + H - A + W - B + 1];
		long[] facInv = new long[W - 2 + H - A + W - B + 1];
		ModInv(W - 2 + H - A + W - B, inv, MOD);
		FacInv(W - 2 + H - A + W - B, inv, fac, facInv, MOD);
		for (int i = 0; i < W - B; i++)
		{
			ans += CombiMod(W - 2 + H - A - i, H - A - 1, MOD, inv, fac, facInv) * CombiMod(i + A - 1, i, MOD, inv, fac, facInv);
			ans %= MOD;
		}
		Console.WriteLine(ans);
	}
	long CombiMod(int n, int k, int mod, long[] inv, long[] fac, long[] facInv)
	{
		if (n == 0 && k == 0) return 1;
		if (n <= 0 || k < 0 || k > n) return 0;
		if (k == 0) return 1;
		
		return ((((long)fac[n] * facInv[k]) % mod) * facInv[n - k]) % mod;
	}

	// モジュラ逆数を求める(modが素数と仮定)
	void ModInv(int n, long[] inv, int mod)
	{
		inv[0] = 0;
		inv[1] = 1;
		for (int i = 2; i <= n; i++)
		{
			inv[i] = mod - ((long)(mod / i) * inv[mod % i] % mod);
		}
	}

	// n!をmodで割った余り
	void FacInv(int n, long[] inv, long[] fac, long[] facInv, int mod)
	{
		fac[0] = facInv[0] = 1;
		for (int i = 1; i <= n; i++)
		{
			fac[i] = (fac[i - 1] * i) % mod;
			facInv[i] = (facInv[i - 1] * (int)inv[i]) % mod;
		}
	}
}

class Scanner
{
	string[] s;
	int i;

	char[] cs = new char[] { ' ' };

	public Scanner()
	{
		s = new string[0];
		i = 0;
	}

	public string next()
	{
		if (i < s.Length) return s[i++];
		string st = Console.ReadLine();
		while (st == "") st = Console.ReadLine();
		s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
		i = 0;
		return next();
	}

	public int nextInt()
	{
		return int.Parse(next());
	}

	public long nextLong()
	{
		return long.Parse(next());
	}

	public double nextDouble()
	{
		return double.Parse(next());
	}
}