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: Time Limit Exceeded)

#include<stdio.h>

int combination(int n, int k)
{
    if (k == 0 || k == n)
        return (1);
    else if (k == 1)
        return (n);
    return (combination(n - 1, k - 1) + combination(n - 1, k));
}

int main(void)
{
    int n, a, b;
	int ans = 0; 

	scanf("%d %d %d", &n, &a, &b);

	if(n%2==1){

	for (int d = 1; d < (n/2)+1;d++){
		ans += combination(n, d)%1000000007;
	}
      
	ans *= 2;
    ans+=1;  
	ans -= combination(n, a)%1000000007;
	ans -= combination(n, b)%1000000007;
	}else{
	for (int d = 1; d < (n/2);d++){
      ans +=combination(n,d)%1000000007;
    }
	ans *= 2;
	ans += (combination(n, n / 2)%1000000007)+1;
	
      ans -= combination(n, a)%1000000007;
	ans -= combination(n, b)%1000000007;	
	}

		printf("%d", ans);
      
	return 0;
}