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: Wrong Answer)

#include<stdio.h>

unsigned int kaijou(unsigned int k){
	int p=k;
	while(p!=1){
		k*=(p-1);
		p--;
	}
	return k;
}
int main(void){
	unsigned int n,sum=0;
	int a,b;
	scanf("%u%d%d",&n,&a,&b );

	for(int i=1;i<=n;i++){

		if(a!=i && b!=i){
			if(i!=0){
			sum+=kaijou(n)/(kaijou((unsigned int)i)*kaijou((unsigned int )i));
			//printf("%u\n",kaijou(n)/(kaijou((unsigned int)i)*kaijou((unsigned int )i)));
			if(n==(kaijou((unsigned int)i)*kaijou((unsigned int )i))){
				sum++;
			}
			//printf(" %u\n",kaijou((unsigned int)i) );
		}
		}
	}
	int k=5;
	printf("%u", sum);
	return 0;
}