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: Wrong Answer)

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

public class GFG
{
    static public void Main()
    {
        try
        {

            var input1 = Console.ReadLine().Split(' ');
            var h = int.Parse(input1[0]);
            var w = int.Parse(input1[1]);
            var a = int.Parse(input1[2]);
            var b = int.Parse(input1[3]);
            //var aar = Console.ReadLine().Split(' ').Select(x => Int32.Parse(x)).ToArray();

            Console.Write(Grid(h, w, a, b));
        }
        catch (Exception ex)
        {
            Console.Write(ex);
        }
    }

    static int Grid(int h, int w, int a, int b)
    {
        int[,] dp = new int[h, w];

        for (int i = 0; i < h; i++)
        {
            dp[i, 0] = 1;
        }

        for (int i = 0; i < w; i++)
        {
            dp[0, i] = 1;
        }

        for(int i=h-a;i<h;i++)
        {
            for(int j=0;j<b;j++)
            {
                dp[i, j] = 0;
            }
        }

        for (int i = 1; i < h; i++)
        {
            for (int j = 1; j < w; j++)
            {
                if (i >= (h - a) && j < b)
                {

                }
                else dp[i, j] = dp[i - 1, j] + dp[i, j - 1];

            }
        }

        return dp[h - 1, w - 1];
    }

}