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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;

public class Main {

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

		String str[] = br.readLine().split(" ");

		long n = Long.parseLong(str[0]);
		long a = Long.parseLong(str[1]);
		long b = Long.parseLong(str[2]);

		long c = modpow(2, n, 1000000007) - 1;

		c -= (c(n, a, 1000000007) % 1000000007 + c(n, b, 1000000007) % 1000000007) % 1000000007;

		System.out.println(c % 1000000007);
	}

	static long c(long n, long k, long mod) {
		long p = 1;
		long q = 1;
		for (long i = 0; i < k; i++) {
			p = p * (n - i) % mod;
			q = q * (i + 1) % mod;
		}
		return p * modpow(q, mod - 2, mod) % mod;
	}

	static long modpow(long a, long n, long mod) {
		long res = 1;
		while (n > 0) {
			if ((n & 1) == 1)
				res = res * a % mod;
			a = a * a % mod;
			n >>= 1;
		}
		return res;
	}

}