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.Numerics;
using System.Linq;
using System.Collections.Generic;
class ProgramD
    {
        static void Main(string[] args)
        {
            
            //入力
            string[] input = Console.ReadLine().Split(' ');
            long n = long.Parse(input[0]);
            long a = long.Parse(input[1]);
            long b = long.Parse(input[2]);

            long MOD = 1000000000 + 7;

            
            
            //まずは総和を計算
            BigInteger temppow = System.Numerics.BigInteger.ModPow(2, n, MOD);
            long ans = (long)temppow;
            if(n/2 < a)
                 a = n - a;
            
            if(n/2 < b)
                 b = n - b;

            long a_sum = 1;
            long b_sum = 1;
            //分子の方をかける
            for(long i = 0;i < a;i++)
            {
                    a_sum *= n - i;
                    a_sum %= MOD;
            }

            //分母の方を割る
            for(long i = 0;i < a;i++)
            {
                a_sum = a_sum * modinv(i + 1,MOD) % MOD;
            }

            //分子の方をかける
            for(long i = 0;i < b;i++)
            {
                    b_sum *= n - i;
                    b_sum %= MOD;
            }

            //分母の方を割る
            for(long i = 0;i < b;i++)
            {
                b_sum = b_sum * modinv(i + 1,MOD) % MOD;
            }

            ans -= a_sum;
            ans %= MOD;
            ans -= b_sum;
            ans -= 1;
            
            if(ans < 0)
                ans = MOD - Math.Abs(ans);
           
            Console.WriteLine(ans);

        }

        // mod. m での a の逆元 a^{-1} を計算する
        static long modinv(long a, long m)
         {
            long b = m, u = 1, v = 0;
            while (b > 0)
            {
                //こちらは普通のユークリッドの互除法
                long t = a / b;
                a -= t * b; 
                var temp = a;
                a = b;
                b = temp;
                //こっちは逆算すると解が出る
                u -= t * v; 
                var temp2 = u;
                u = v;
                v = temp2;
            }
            u %= m;
            if (u < 0) 
                u += m;
            return u;
        }
            
    }