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: Accepted)

char buf[99];sz;
int N,A,B;
long mod=1e9+7;
long p(long a,long b){return b?p(a*a%mod,b>>1)*(b&1?a:1)%mod:1;}
long C(long a,long b)
{
	long x=1,y=1;
	for(int i=1;i<=b;i++)
	{
		x=x*(a-i+1)%mod;
		y=y*i%mod;
	}
	return x*p(y,mod-2)%mod;
}
main()
{
	read(0,buf,99);
	for(;;)
	{
		char c=buf[sz++];
		if(c<'0')break;
		N=N*10+c-'0';
	}
	for(;;)
	{
		char c=buf[sz++];
		if(c<'0')break;
		A=A*10+c-'0';
	}
	for(;;)
	{
		char c=buf[sz++];
		if(c<'0')break;
		B=B*10+c-'0';
	}
	printf("%d\n",((p(2,N)-1-C(N,A)-C(N,B))%mod+mod)%mod);
}