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: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>
#define MOD 1000000007

long	factorial(long nb)
{
	long	i;
	long	factorial;

	if (nb < 0)
		return (0);
	factorial = 1;
	i = 2;
	while (i <= nb)
		factorial = (factorial * i++) % MOD;
	return (factorial);
}

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

	i = m - n + 1;
	comb = 1;
	while (i <= m)
		comb = (comb * i++) % MOD;
	comb /= factorial(n);
	return (comb);
}

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

	if ((scanf("%d %d %d", &n, &a, &b)) < 3)
		return (-1);
	ans = 0;
	i = 1;
	while (i <= n)
	{
		if (i != a && i != b)
			ans = (ans + combination(n, i)) % MOD;
		i++;
	}
	printf("%ld", ans);
}