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 {
    static long[] fac = new long[1000003];
    static long[] finv = new long[1000003];
    static long[] inv = new long[1000003];
    static final int MOD = 1000000007;
    public static void main(String[] args) {
        @SuppressWarnings("resource")
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int a  = sc.nextInt();
        int b = sc.nextInt();

        long aa = 1;
        long bb = 1;
        //前処理
        fac[0] = fac[1] = 1;
        finv[0] = finv[1] = 1;
        inv[1] = 1;
        for (int i = 2; i < 1000003; i++) {
            fac[i] = fac[i - 1] * i % MOD;
            inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
            finv[i] = finv[i - 1] * inv[i] % MOD;
        }
        for (int i = 0 ; i < a ; i++) {
            aa *= (long) (n - i) % MOD;
            aa %= MOD;
        }
        for (int i = 0 ; i < b ; i++) {
            bb *= (long) (n - i) % MOD;
            bb %= MOD;
        }

        aa *= finv[a] % MOD;
        bb *= finv[b] % MOD;
        aa %= MOD;
        bb %= MOD;
        long ans = (pow (2, n) - 1 - aa - bb) % MOD;
        if (n == 2) {
            System.out.println(0);
            return;
        }
        System.out.println((ans + MOD) % MOD);
    }

    static int pow (long b, int e) {
        long ans = 1;
        while (e > 0) {
            if (e % 2 == 1) {
                ans = ans * b % MOD;
            }
            e >>= 1;
            b = b * b % MOD;
        }
        return (int) ans;
    }

}