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.*;

public class Main {

	static long mod = 1_000_000_007;
	
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		long n = scan.nextInt();
		long a = scan.nextInt();
		long b = scan.nextInt();
		long nt = pow(2, n);
		nt -= nck(n,a);
		nt %= mod;
		if(nt < 0) nt += mod;
		nt -= nck(n, b);
		nt %= mod;
		if(nt < 0) nt += mod;
		nt--;
		nt %= mod;
		if(nt < 0) nt += mod;
		System.out.println(nt);
	}
	
	static long nck(long n, long k) {
		// small k
		long kfac = 1;
		for(int i = 1; i <= k; i++){
			kfac *= i;
			kfac %= mod;
		}
		long num = 1;
		for(long i = n; i > n-k; i--) {
			num *= i;
			num %= mod;
		}
//		System.out.println(num+" "+kfac);
		long minv = pow(kfac, mod-2);
//		System.out.println(num*minv%mod);
		return (num*minv)%mod;
	}
	
	static long pow(long x, long b) {
		if(b == 0) return 1;
		if(b % 2 == 1) {
			long r = pow(x, b-1);
			return (r*x)%mod;
		}
		long r = pow(x, b/2);
		return (r*r)%mod;
	}

}