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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;

public class AtCoderD
{
    static void Main()
    {
        int n;
        int a;
        int b;
        {
            var line = Console.ReadLine();
            var words = line.Split(' ');
            n = int.Parse(words[0]);
            a = int.Parse(words[1]);
            b = int.Parse(words[2]);
        }

        int mod = 1000000007;
        long total = Pow(2, n, mod);
        long ca = Combination(n, a, mod);
        long cb = Combination(n, b, mod);
        long answer = total - ca - cb - 1;

        answer += mod * 2;
        answer = answer % mod;

        Console.WriteLine(answer);
    }

    // Comnibation（mod版）
    private static long Combination(int n, int k, int d)
    {
        long answer = 1;
        for (int i = n; i > n - k; i--)
        {
            answer = Mul(answer, i, d);
        }
        for (int i = 1; i <= k; i++)
        {
            answer = Div(answer, i, d);
        }
        return answer;
    }

    // 掛け算（mod版）
    private static long Mul(long a, long b, long c)
    {
        return ((a % c) * (b % c)) % c;
    }

    // 割り算（mod版）
    private static long Div(long a, long b, long c)
    {
        return Mul(a, Pow(b, c - 2, c), c);
    }

    // べき乗（mod版）
    private static long Pow(long a, long b, long c)
    {
        if (b == 0)
        {
            return 1;
        }
        else if (b == 1)
        {
            return a % c;
        }
        else if (b % 2 == 0)
        {
            long d = Pow(a, b / 2, c);
            return d * d % c;
        }
        else
        {
            return a * Pow(a, b - 1, c) % c;
        }
    }
}