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)

#include<stdio.h>

#define MODNUM 1000000007

long long modpow(long long a, long long n, long long mod){
	long long result = 1;
	while(n > 0){
		if(n & 1){
			result = result * a % mod;
		}
		a = a * a % mod;
		n /= 2;
	}

	return result;
}

long long modinv(long long a, long long mod){
	return modpow(a, mod - 2, mod);
}


int main(){
	int n, a, b;
	long long ans = 0;
	int i;
	long long m, c;

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

	ans = (modpow(2, n, MODNUM) - 1) % MODNUM;

	m = 1;
	c = n;
	for(i=2; i<=a; i++){
		m = m * i % MODNUM;
		c = c * (n-i+1) % MODNUM;
	}
	ans = (ans - c * modinv(m, MODNUM)) % MODNUM;
	for(; i<=b; i++){
		m = m * i % MODNUM;
		c = c * (n-i+1) % MODNUM;
	}
	ans = (ans - c * modinv(m, MODNUM)) % MODNUM;

	while(ans < 0){
		ans += MODNUM;
	}

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