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 <algorithm>
#include <iomanip>
#include <iostream>
#include <numeric>
#include <vector>

const long mod = 1000000000 + 7;

long mod_pow(long a, long x, long m)
{
  if (x <= 0) return 1;
  return x & 1
      ? a * mod_pow(a, x - 1, m) % m
      : mod_pow(a * a % m, x >> 1, m);
}

long mod_combi(long n, long k, long m)
{
  long nume = 1;
  long deno = 1;
  for (long i = 0; i < k; ++i) {
    nume = nume * (n - i) % m;
    deno = deno * (k - i) % m;
  }
  return nume * mod_pow(deno, m - 2, m) % m;
}

long mod_sub(long a, long b, long m)
{
  const auto r = (a - b) % m;
  return r >= 0 ? r : r + m;
}

long solve(long n, long a, long b)
{
  return mod_sub(
      mod_sub(
        mod_sub(mod_pow(2, n, mod), 1, mod),
        mod_combi(n, a, mod),
        mod),
      mod_combi(n, b, mod),
      mod);
}

int main()
{
  long n, a, b;
  std::cin >> n >> a >> b;

  std::cout << std::setprecision(20);
  std::cout << solve(n, a, b) << std::endl;
}