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>
#include<math.h>

double Combi(int n, int r){

  int i;
  double c = 1.0;  

  if((1 <= r) && (r < n)){
    for(i=r;i>=1;i--){
      c *= ((n + 1.0 - (double)i) / (double)i);
    }
  }
  
  return c;
}

int main(){
  int n,a,b;
  int i;
  int val=0;
  int ans=0;
  int sum=0;
  scanf("%d %d %d",&n,&a,&b);
  
  for(i=1;i<=n;i++){
  val=sum+Combi(n,i);
  sum=sum+val%(int)(pow(10,9)+7);
  }  
    
  
  printf("%d",ans);
  val=Combi(n,a);
  ans=ans-val%(int)(pow(10,9)+7);
  val=Combi(n,b);
  ans=ans-val%(int)(pow(10,9)+7);
  
  printf("%d",ans);
  return 0;
}