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: Accepted)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Main main = new Main();
		main.run();
	}

	final static int MOD = 1000000007;
	public void run() {
		Scanner sc = new Scanner(System.in);
		long n = sc.nextLong();
		long a = sc.nextLong();
		long b = sc.nextLong();

		long ans = pow(2, n);
		ans = minus(ans, 1);
		ans = minus(ans, combi(n, a));
		ans = minus(ans, combi(n, b));
		System.out.println(ans);
		sc.close();
	}

	private long combi(long n, long k) {
		long x=1;
		long y=1;
		for(int i=0; i<k; i++) {
			x = mul(x, (n-i));
			y = mul(y, (i+1));
		}
		return div(x, y);
	}

	private long div(long a, long b) {
		return mul(a, inv(b));
	}

	private long inv(long n) {
		return pow(n, MOD-2);
	}

	private long pow(long n, long x) {
		long res = 1;
		if(x==0) return res;
		res = pow(n, x/2);
		if(x%2==0) {
			res = mul(res, res);
		}else {
			res = mul(mul(res, res), n);
		}
		return res;
	}

	private long minus(long a, long b) {
		if((a -= b) < 0) a += MOD;
		return a;
	}

	private long mul(long a, long b) {
		return (a * b) % MOD;
	}
}