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: Accepted)

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



class Myon
{
    public Myon() { }
    public static int Main()
    {
        new Myon().calc();
        return 0;
    }


    Scanner cin;



    void calc()
    {
        cin = new Scanner();
        setFact(250000);
        int H = cin.nextInt();
        int W = cin.nextInt();
        long ans = getC(H - 1 + W - 1, W - 1);
        int A = cin.nextInt();
        int B = cin.nextInt();
        for (int i = 0; i < B; i++)
        {
            ans -= getC(H - 1 - A + i, i) * getC(A - 1 + W - 1 - i, W - 1 - i);
            ans %= mod;
            ans += mod;
            ans %= mod;
        }
        Console.WriteLine(ans);
    }

    long mod = 1000000007;
    long powmod(long a, long p)
    {
        long ans = 1;
        long mul = a;
        for (; p > 0; p >>= 1, mul = (mul * mul) % mod)
        {
            if ((p & 1) == 1) ans = (ans * mul) % mod;
        }
        return ans;
    }


    long[] fact; //階乗
    long[] revFact; //階乗の逆元
    void setFact(int N)
    {
        fact = new long[N];
        revFact = new long[N];
        fact[0] = 1;
        for (int i = 1; i < N; i++)
        {
            fact[i] = fact[i - 1] * i;
            fact[i] %= mod;
        }
        revFact[N - 1] = powmod(fact[N - 1], mod - 2);
        for (int i = N - 2; i >= 0; i--)
        {
            revFact[i] = revFact[i + 1] * (i + 1);
            revFact[i] %= mod;
        }
    }


    long getC(int a, int b)
    {
        return (((fact[a] * revFact[b]) % mod) * revFact[a - b]) % 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);
        if (s.Length == 0) return next();
        i = 0;
        return s[i++];
    }

    public int nextInt()
    {
        return int.Parse(next());
    }
    public int[] ArrayInt(int N, int add = 0)
    {
        int[] Array = new int[N];
        for (int i = 0; i < N; i++)
        {
            Array[i] = nextInt() + add;
        }
        return Array;
    }

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

    public long[] ArrayLong(int N, long add = 0)
    {
        long[] Array = new long[N];
        for (int i = 0; i < N; i++)
        {
            Array[i] = nextLong() + add;
        }
        return Array;
    }

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


    public double[] ArrayDouble(int N, double add = 0)
    {
        double[] Array = new double[N];
        for (int i = 0; i < N; i++)
        {
            Array[i] = nextDouble() + add;
        }
        return Array;
    }
}