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 Mint
  MOD = 1_000_000_007

  attr_reader :x

  def initialize(x)
    @x = ((x % MOD) + MOD) % MOD
  end

  def to_s
    @x.to_s
  end

  def -@
    Mint.new(-@x)
  end

  def +(a)
    Mint.new(@x).add(a)
  end

  def -(a)
    Mint.new(@x).sub(a)
  end

  def *(a)
    Mint.new(@x).mul(a)
  end

  def /(a)
    Mint.new(@x).div(a)
  end

  def add(a)
    @x += a.x
    @x -= MOD if @x >= MOD
    self
  end

  def sub(a)
    @x += MOD - a.x
    @x -= MOD if @x >= MOD
    self
  end

  def mul(a)
    @x = (@x * a.x) % MOD
    self
  end

  def pow(t)
    return Mint.new(1) if t.zero?
    a = pow(t >> 1)
    a = a.mul(a)
    a = a.mul(self) if (t % 2) == 1
    a
  end

  def inv
    pow(MOD - 2)
  end

  def div(a)
    self.mul(a.inv)
  end

  def factorial
    return Mint.new(1) if @x.zero?
    (1..@x).map{|v| Mint.new(v)}.inject(:mul)
  end

  def permutation(k)
    return Mint.new(1) if k.zero?
    (@x - k + 1..@x).map{|v| Mint.new(v)}.inject(:mul)
  end

  def combination(k)
    permutation(k).div(Mint.new(k).factorial)
  end
end

N, A, B = gets.chomp.split(" ").map(&:to_i)
a = [A, N-A].min
b = [B, N-B].min
n = Mint.new(N)

puts Mint.new(2).pow(N) - n.combination(a) - n.combination(b) - Mint.new(1)