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: Wrong Answer)

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


class Program
{
    static long[] kaijo;
    static long[] Ikaijo;
    static long[] kaijon;
    static long mo = 1000000007;
    
    static void Main(string[] args)
    {
        //入力を受け取る
        var NAB = Console.ReadLine().Split(' ').Select(x => long.Parse(x)).ToArray();
        long N = NAB[0];
        long A = NAB[1];
        long B = NAB[2];
        //N個のものからR個選ぶ組み合わせ
        kaijo = new long[B + 1];
        //階乗のモジュラー逆数
        Ikaijo = new long[B + 1];
        kaijon = new long[B + 1];
        kaijo[0] = 1;
        Ikaijo[0] = 1;
        kaijon[0] = N;
        
        for (long i = 1; i < B + 1; i++)
        {
            kaijo[i] = (kaijo[i - 1] * i) % mo;
            Ikaijo[i] = RepeatSquaring(kaijo[i]);
            kaijon[i] = (kaijon[i - 1] * (N - i)) % mo;
        }
        
        long ans = RepeatSquaring(2, N) - (nCr(B) + nCr(A))%mo;
        if(A !=1){
            ans--;
        }
        Console.WriteLine(ans%mo);
    }
    
    static long nCr(long r){
        return ((kaijon[r-1]%mo) * (Ikaijo[r]%mo))%mo;
    }


    //繰り返し2乗法
    static long RepeatSquaring(long N, long M = 1000000005, long mod = 1000000007)
    {
        if (M == 0)
        {
            return 1;
        }
        else if (M % 2 == 0)
        {
            long t = RepeatSquaring(N, (M / 2), mod) % mod;
            return (t * t) % mod;
        }
        else
        {
            return ((N % mod) * RepeatSquaring(N, M - 1, mod)) % mod;
        }
    }
}