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.Linq;//リストの使用
using System.Collections.Generic;
using System.Text;//テキストの高速出力に必要
class Program
{
static long n = 210000;
static long[] factorials = new long[n+1];//i!(mod p)を先にメモ
static long[] factorialRs = new long[n+1];//i!^-1(mod p), pは素数

	static void Main()
	{
		string[] input = Console.ReadLine().Split(' ');//Splitで区切り文字を指定して複数個受け取る。
		long h = long.Parse(input[0]);
		long w = long.Parse(input[1]);
    long a = long.Parse(input[2]);
		long b = long.Parse(input[3]);
		long p = 1000000007;
    long answerm = 0;
    long answer = 0;

    factorials[0] = 1;
		factorialRs[n] = DivideModFactorial(n,p);
		for(long i = 1; i <= n; i++)
		{
		factorials[i] = (factorials[i-1]*i)%p;//i!(mod p)
		factorialRs[n-i] = (factorialRs[n+1-i]*(n+1-i))%p;//逆元も先にメモ
		}

    for(int i = 0; i < h-a; i++)//下に行く回数
    {
      answer = factorials[b-1+i];
      answer %= p;
      answer *= factorialRs[b-1];
      answer %= p;
      answer *= factorialRs[i];
      answer %= p;

      answer *= factorials[w-b-1 + h-i-1];
      answer %= p;
      answer *= factorialRs[w-b-1];
      answer %= p;
      answer *= factorialRs[h-i-1];
      answer %= p;

      //Console.WriteLine(answer);
      answerm += answer;
      answerm %= p;
    }

		Console.WriteLine(answerm);
	}

  static long DivideMod(long x, long a, long p)//戻り値はx^a(mod p)
  {
    long num = 2;
    long answer = 1;
    long check = a;
    long memo = x%p;
    
    while(check > 0)
    {
      if(check % num == num / 2)
      {
        check -= num / 2;
        answer *= memo;
        answer %= p;
      }
    memo *= memo;
    memo %= p;
    num *= 2;
    }
    return answer;
  }

  static long DivideModReverse(long x, long p)//戻り値はx^-1(mod p), pは素数
  {
    long answer = DivideMod(x, p-2, p);
    return answer;
  }

  static long DivideModFactorial(long x, long p)//戻り値はx!^-1(mod p), pは素数
  {
    long answer = 1;
    for(long i = x; i >= 2; i--)
    {
      answer *= DivideModReverse(i, p);
      answer %= p;
    }
    return answer;
  }

  static long Comb(long a, long b, long p)//戻り値は組み合わせC(a,b)のmod p
  {
    long answer = 1;
    answer *= factorials[a];
    answer %= p;
    answer *= factorialRs[a-b];//引数a-bは負にならないようにする
    answer %= p;
    answer *= factorialRs[b];
    answer %= p;
    return answer;
  }

  
}