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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

/*
 * 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 10^9+7 で割ったあまりを出力してください。
 */

import java.util.*;
	public class Main{
		public static void main(String[] args){
            Scanner sc = new Scanner(System.in);
            // 入力
            int h = Integer.parseInt(sc.next());
            int w = Integer.parseInt(sc.next());
            int a = Integer.parseInt(sc.next());
            int b = Integer.parseInt(sc.next());
            
            int[] dp = new int[w];
            
            for(int i=0; i<w; i++){
            	dp[i] = 1;
            }
            
            for(int i=1; i<h-a; i++){
            	for(int j=1; j<w; j++){
            		dp[j] = (dp[j] + dp[j-1]) % (int)(Math.pow(10, 9) + 7);
            	}
            }
            
            for(int i=h-a; i<h; i++){
            	for(int j=b+1; j<w; j++){
            		dp[j] = (dp[j] + dp[j-1]) % (int)(Math.pow(10, 9) + 7);
            	}
            }
            
            // 出力
            System.out.println(dp[w-1]);

            sc.close();
		}
	}