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>
#define MOD 1000000007

long	my_pow(long  x, long n)
{
	if (n == 0)
		return (1);
	if (n % 2 == 0)
		return ((my_pow((x * x) % MOD, n / 2)) % MOD);
	if (n % 2 == 1)
		return ((my_pow((x * x) % MOD, n / 2) * x) % MOD);
	return (1);
}

int		factorial(int n)
{
	if (n < 0)
		return (0);
	if (n < 1)
		return (1);
	return ((int)(((long)n * factorial(n - 1)) % MOD));
}

int		inverse(int nb)
{
	return ((int)(my_pow(nb, MOD - 2) % MOD));
}

int		comb(int m, int n)
{
	int		i;
	long	comb;

	comb = 1;
	i = m - n + 1;
	while (i <= m)
	{
		comb *= i++;
		comb %= MOD;
	}
	printf("comb%ld\n", comb);
	printf("fact%d\n", factorial(n));
	comb *= inverse(factorial(n));
	comb %= MOD;
	printf("inv%ld\n", comb);
	return ((int)comb);
}

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

	if ((scanf("%d %d %d", &n, &a, &b)) < 3)
		return (-1);
	printf("%d\n", (int)my_pow(2, (long)n));
	printf("%d\n", comb(n, a));
	printf("%d\n", comb(n, b));
	ans = (((int)my_pow(2, (long)n) - 1 - comb(n, a)) % MOD - comb(n, b)) % MOD;
	if (ans < 0)
		ans += MOD;
	printf("%d", ans);
}