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)

MOD = 10 ** 9 + 7

# MODを法とした階乗、組み合わせ
class ModCombination
  attr_reader :fac, :inv, :finv, :mod
 
  def initialize(n, mod: MOD)
    @mod = mod
    @fac = [1, 1]
    @inv = [1, 1]
    @finv = [1, 1]
 
    (2..n).each do |i|
      @fac[i] = fac[i - 1] * i % mod
      @inv[i] = mod - inv[mod % i] * (mod / i) % mod
      @finv[i] = finv[i - 1] * inv[i] % mod
    end
  end
 
  def combination(n, k)
    return 0 if n < k
    return 0 if n < 0 || k < 0
 
    fac[n] * (finv[k] * finv[n - k] % mod) % mod
  end
  
  def repeated_combination(n, k)
    combination(n + k - 1, k)
  end
end

# MODを法としたxのy乗を求める（繰り返し二乗法）
def modPow(x,y)
  ans = 1
  while y > 0
    ans = ans * x % MOD if y & 1 == 1
    y /= 2
    x = x * x % MOD
  end
  return ans
end

MC = ModCombination.new(300000)

# MODを法としたnCk、kが小さいとき
def nCk(n,k)
  ans = n
  2.upto(k) do |i|
    ans = ans * (n - i + 1) % MOD
    ans = ans * MC.inv[i] % MOD
  end
  ans
end

n,a,b = gets.split.map &:to_i
x = modPow(2,n)
y = MOD - nCk(n,a)
z = MOD - nCk(n,b)
puts (x + y + z - 1) % MOD