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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

#define F first
#define S second

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;

const int N = 15, M = 10, mod = 1e9 + 7;

int n, a, b;

int qp(int a, int b)
{
	int res = 1 % mod;
	while (b)
	{
		if (b & 1) res = (LL)res * a % mod;
		a = (LL)a * a % mod;
		b >>= 1;
	}
	return res;
}

int C(int a, int b)
{
	int res = 1;
	for (int i = a; i > a - b; i -- ) res = (LL)res * i % mod;
	for (int i = 1; i <= b; i ++ ) res = (LL)res * qp(i, mod - 2) % mod;
	return res;
}

int main()
{
	cin >> n >> a >> b;
	
	int res = (qp(2, n) - 1 + mod) % mod;
	res = (res - C(n, a) + mod) % mod;
	res = (res - C(n, b) + mod) % mod;

	cout << res << endl;
	return 0;
}