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 long MOD = (long) 1e9 + 7;

    public static void main(final String[] args) {
        final Scanner scanner = new Scanner(System.in);
        final long n = scanner.nextInt();
        final long a = scanner.nextInt();
        final long b = scanner.nextInt();

        long all = 1;
        for (int i = 10; i <= n; i += 10) {
            all *= 1024;
            all %= MOD;
        }
        for (int i = 1; i <= n % 10; i++) {
            all *= 2;
            all %= MOD;
        }

        all--;

        System.out.println((all + MOD - calc(n, a) + MOD - calc(n, b)) % MOD);
    }

    private static long calc(final long n, final long k) {
        long tmp = 1;
        for (long i = 1; i <= k; i++) {
            tmp *= ((n - i + 1) * modInv(i, MOD)) % MOD;
            tmp %= MOD;
        }
        return tmp;
    }

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

    private static long modInv(final long a, final long mod) {
        return modPow(a, mod - 2, mod);
    }
}