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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

int64_t comb(int64_t n, int64_t r){
  if(r==1){
    return n;
  }
  else if(r==n-1){
    return n;
  }
  else{
    return comb(n-1,r-1) + comb(n-1,r);
  }
}

int64_t comb2(int64_t n, int64_t r){
  int64_t result = 1;
  for(int i=1; i<=r; i++){
    result = result * (n-r+i) / i;
  }
  return result;
}

int main() {
  int64_t n, a, b;
  cin >> n >> a >> b;
  int64_t result = 1;
  for (int i=0; i<n; i++){
    result = result*2;
    if(result>=pow(10,9)+7){
      result-=pow(10,9)+7;
    }
  }
  result = result - comb2(n,a) - comb2(n,b) - 1;
  while((result<0) || (result>=pow(10,9)+7)){
    if(result<0){
      result+=pow(10,9)+7;
    }
    else if(result>=pow(10,9)+7){
      result-=pow(10,9)+7;
    }
  }
  cout << result << endl;
}