Problem:
Score : 400 points

Problem Statement

Akari has n kinds of flowers, one of each kind.

She is going to choose one or more of these flowers to make a bouquet.

However, she hates two numbers a and b, so the number of flowers in the bouquet cannot be a or b.

How many different bouquets are there that Akari can make?

Find the count modulo (10^9 + 7).

Here, two bouquets are considered different when there is a flower that is used in one of the bouquets but not in the other bouquet.

Constraints

All values in input are integers.

2 \leq n \leq 10^9

1 \leq a < b \leq \textrm{min}(n, 2 \times 10^5)

Input

Input is given from Standard Input in the following format:

n a b

Output

Print the number of bouquets that Akari can make, modulo (10^9 + 7). (If there are no such bouquets, print 0.)

Sample Input 1

4 1 3

Sample Output 1

7

In this case, Akari can choose 2 or 4 flowers to make the bouquet.

There are 6 ways to choose 2 out of the 4 flowers, and 1 way to choose 4, so there are a total of 7 different bouquets that Akari can make.

Sample Input 2

1000000000 141421 173205

Sample Output 2

34076506

Print the count modulo (10^9 + 7).

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

import java.math.BigInteger;
import java.util.Scanner;

public class Main
{
	static public BigInteger mod = new BigInteger("1000000007");
	public static void main(String[] args)
	{
		//int m = 1000000007;

		BigInteger ans = BigInteger.ZERO;
		Scanner scan = new Scanner(System.in);
		String str = scan.nextLine();
		String[] substr = str.split(" ", 0);

		int n = Integer.parseInt(substr[0]);
		int a = Integer.parseInt(substr[1]);
		int b = Integer.parseInt(substr[2]);

		for (int i = 1; i < n+1; i++) {
			if (i != a && i != b) {
				ans = ans.add(f(n, i));
			}
		}
		scan.close();
		System.out.println(ans);
	}

	static public BigInteger f(int n, int m)
	{
		BigInteger ans = BigInteger.ONE;
		int k = n;
		for (int i = 0; i < m; i++) {
			ans = ans.multiply(BigInteger.valueOf(k));
			k--;
		}
		BigInteger ans2 = BigInteger.ONE;
		for (int i = 1; i < m+1; i++) {
			ans2 = ans2.multiply(BigInteger.valueOf(i));
		}
		ans = ans.divide(ans2);
		ans = ans.remainder(mod);
		return ans;
	}
}