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>
#define rep(i,n) for(int i=0; i<(n); ++i)
#define all(x) (x).begin(),(x).end()
#define pb push_back
using namespace std;
typedef long long ll;

long nCr(int n, int r) {
    long ans = 1;
    for (int i = n; i > n - r; --i) {
        ans = ans*i;
    }
    for (int i = 1 ; i < r + 1; ++i) {
        ans = ans / i;
    }
    return ans;
}

int main()
{
	int n,a,b;
	ll sum=0;
	scanf("%d %d %d",&n,&a,&b);
  
  	for(int i=0; i<=n/2; i++){
//      if(i==a || i==b) continue;
      sum += nCr(n,i);
    }
  	cout << sum << endl;
  	sum = sum % (1000000007);
  	sum *= 2;
  	sum = sum - 1 - nCr(n,a) - nCr(n,b);
  	if(n%2 == 0) sum -= nCr(n,n/2);
  	sum = sum % (1000000007);
	cout << sum << endl;
  	return 0;
}