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 {
    private static final int MOD = 1_000_000_007;
    private static final String YES = "Yes";
    private static final String NO = "No";

    private static int[] fact;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int N = sc.nextInt();
        int a = sc.nextInt();
        int b = sc.nextInt();

        System.out.println(solve(N, a, b));
    }

    private static long solve(int N, int a, int b) {
        long sum = power(2, N, MOD) - 1;

        sum = mod(sum - combination(N, a, MOD), MOD);
        sum = mod(sum - combination(N, b, MOD), MOD);

        return mod(sum, MOD);
    }

    public static long power(long base, long t, long p) {
        long ans = 1;

        while (t > 0) {
            if (t % 2 == 1) {
                ans = ans * base % p;
            }

            t /= 2;
            base = base * base % p;
        }

        return ans;
    }

    public static long combination(int n, int r, int p) {
        if (n == 0 && r == 0) return 1;
        if (n < r || n < 0)   return 0;

        r = Math.min(r, n-r);
        long x = 1;
        long y = 1;
        for (int i=1; i<=r; i++) {
            x = x * (n-i+1) % p;
            y = y * i % p;
        }

        return mod(x * power(y, p-2, p), p);
    }

   public static long mod(long a, long p) {
        return (a%p + p) % p;
    }
}