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 Ruby (2.3.3) (judge verdict: Accepted)

class Integer
  attr_accessor :mod
  @mod = nil
  def self.set_mod(mod = 10**9 + 7)
    @mod = mod
  end

  def combination(k = self) # 4C2 = 6
    return 1 if k.zero?
    if @mod.nil?
      (self - k + 1..self).inject(:*) / k.factorial
    else
      (self - k + 1..self).inject{|i,m|i * m % @mod} * k.factorial.modinv % @mod
    end
  end

  def permutation(k = self) # 4P2 = 12
    return 1 if k.zero?
    if @mod.nil?
      (self - k + 1..self).inject(:*)
    else
      (self - k + 1..self).inject{|i,m|i * m % @mod}
    end
  end

  def factorial(step = 1) # 4! = 24
    return 1 if self.zero?
    if @mod.nil?
      (1..self).inject(:*)
    else
      (1..self).inject{|i,m|i * m % @mod}
    end
  end

  def modinv # https://qiita.com/drken/items/3b4fdf0a78e7a138cd9a
    mod = 10**9 + 7
    a = self; b = mod; u = 1; v = 0
    while b > 0
      t = a / b
      a -= t * b
      a, b = b, a
      u -= t * v
      u, v = v, u
    end
    u %= mod
    u += mod if u < 0
    u
  end

  # 繰り返し二乗法
  def repeat_squaring(p, mod = (10**9 + 7))
    return 1 if p == 0
    if p % 2 == 0
      t = repeat_squaring(p/2, mod)
      return t*t % mod
    end
    self * repeat_squaring(p - 1, mod) % mod
  end

  alias :C :combination
  alias :P :permutation
  alias :! :factorial
end

Integer.set_mod
n, a, b = gets.split.map(&:to_i)
mod = 10 ** 9 + 7


def nCk(n,k)
  mod = 10 ** 9 + 7
  (1..k).map{|x| n - x + 1}.inject{|m,i|i*m % mod} * (1..k).inject{|m,i|i*m % mod}.modinv
end

# x = (2.repeat_squaring(n) % mod) - n.C(a) - n.C(b) - 1
x = (2.repeat_squaring(n) % mod) - nCk(n,a) - nCk(n,b) - 1


p x % mod
# 7 34076506