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: Wrong Answer)

import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		try (Scanner sc = new Scanner(System.in)) {
			int n = sc.nextInt();
			int a = sc.nextInt();
			int b = sc.nextInt();
			int div = 1000000007;
			long ans = pow(2, n, div);
			int A = nCk(n, a, div);
			int B = nCk(n, b, div);
			ans -= A;
			ans -= B;
			ans -= 1;
			if (ans < 0) {
				System.out.print((div + ans) % div);
			} else {
				System.out.print(ans % div);
			}
		}
	}

	public static int nCk(int n, int k, int M) {
		long ret = 1;
		int min = Math.min(k, n - k);
		for (int i = 1; i <= min; i++) {
			ret = (ret * pow(i, M - 2, M)) % M;
		}
		for (int i = n - min + 1; i <= n; i++) {
			ret = (ret * i) % M;
		}
		return (int) ret;
	}

	//return a^b mod M O(logB)
	public static long pow(long a, long b, int M) {
		long ret = 1;
		long tmp = a;
		while (b > 0) {
			if ((b & 1) == 1)
				ret = (ret * tmp) % M;
			tmp = (tmp * tmp) % M;
			b = b >> 1;
		}
		return ret;
	}
}