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: Runtime Error)

#include<stdio.h>

int n,a,b;
int m_r,fac_r,fac_m,fac_m_r,m;

int fac(int n){
	int ans=1;
	for(;n>=1;n--){
		ans=ans*n;
	}
	return ans;
}

int main(){������
	int total=0;
	int r =1;
	scanf("%d %d %d",&m,&a,&b);
	for(;m>=r;r++){
		fac_m=fac(m);
		m_r=m-r;
		fac_m_r=fac(m_r);
		fac_r=fac(r);
		if (r != a && r != b){
			total+=fac_m/(fac_r*fac_m_r);
		}
		else
		{
		//	printf("_%d\n",r);
		}
		//	printf("%d %d %d %d\n",m,m_r,r,total);
	}
	printf("%d",total);
}