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

  # O(min(n-k,k) log m)
  # combination: nPk / k! (mod p), note: m must be a prime number
  def cmb(k, m=10**9+7)
    n = self
    k = n - k if k > n - k
    n.prm(k, m) * k.prm(k,m).modinv(m) % m
  end

  # mod must be a prime number
  # O(log m)
  def modinv(mod = 10**9+7)
    pow(mod-2, mod)
  end

  # O(log n)
  def pow(n , mod = nil )
    a = self.to_i
    res = 1
    while n > 0
      if (n & 1) != 0
        res *= a
        res %= mod if mod
      end
      a *= a
      a %= mod if mod
      n >>= 1
    end
    res
  end
end


class Facts

  def initialize(mod=10**9+7, n_max=1)
    @mod = mod
    @n_max = n_max
    @fact = [1, 1]
    @inv = [0,1]
    @factinv = [1, 1]
    @perms = Hash.new{|h,k| h[k] = [1] }
    setup_table(n_max) if 1 < @n_max
  end

  def cmb(n,r)
    return 0 if r < 0 || n < r
    r = n - r if r > n - r
    setup_table(r) if @n_max < r
    return prm(n,r) * @factinv[r] % @mod
    # @fact[n] * (@factinv[r] * @factinv[n-r] % @mod) % @mod
  end

  def factorial(n)
    setup_table(n) if @n_max < n
    @fact[n]
  end

  def prm(n,k)
    
    if @perms[n][k].nil?
      perm = @perms[n]
      (n-perm.size+1).downto(n-k){|i| perm.push( perm[-1] * i % @mod )  }
    end
    return @perms[n][k]
    
    setup_table(n) if @n_max < n
    @fact[n] * @factinv[n-k] % @mod
  end

  private
    # O(t)
    def setup_table(t)
      (@n_max+1).upto(t) do |i|
        @fact.push( @fact[-1] * i % @mod )
        @inv.push( -@inv[@mod % i] * (@mod / i) % @mod )
        @factinv.push( @factinv[-1] * @inv[-1] % @mod )
      end
      @n_max = t
    end
end

mod = 10 ** 9 + 7

n,a,b = gets.split.map(&:to_i)

#何も選ばない1通りを除く
f = Facts.new
# ans = 2.pow(n, mod) - 1
# p f.cmb(n,a)
ans = 2.pow(n, mod) - 1 - f.cmb(n,a) - f.cmb(n,b)
ans %= mod
 
puts ans

# p nil.to_i
# p h = Hash.new{|h,k| h[k] = [1] }
# p h[1][4]
# p h[4]