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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#define N 1000000007

long long powtwo(int,int);
long long con(int,int);
int main(){
  int n,a,b,sum; scanf("%d%d%d",&n,&a,&b);
  //printf("%lld\n",con(12,5));
  sum=powtwo(2,n)-1-con(n,a)-con(n,b);
  printf("%lld\n",sum%N);
  return 0;
}

long long powtwo(int k,int n){
  if (n==0) return 1;
  else{
    if(n%2==0) return powtwo(k*k,n/2)%N;
    else return powtwo(k,n-1)*k%N;
  }
}

long long con(int n, int k){
  long long ans; if(k>n/2) return con(n,n-k);
  if (n==0 || k==0 ) ans=1;
  else ans=n*con(n-1,k-1)/k;
  return ans%N;
}