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

def pow(x, n, mod = MOD)
  ans = 1
  while n > 0 do
    ans = ans * x % mod if n.odd?
    x = x * x % mod
    n >>= 1
  end
  return ans
end

def inv(x, mod)
  return pow(x, mod - 2, mod)
end

@fact = [1]
def fact(n, mod = MOD)
  return @fact[n - 1] unless @fact[n - 1].nil?

  f = @fact[-1]
  (@fact.size + 1).upto(n) do |i|
    f = (f * i) % mod
    @fact << f
  end
  return @fact[n - 1]
end

def perm(n, r, mod = MOD)
  return n.downto(n - r + 1).reduce(1) {|ans, x| (ans * x) % mod }
end

def comb(n, r, mod = MOD)
  return 0 unless 0 <= r && r <= n
  return 1 if r == 0 || r == n
  return perm(n, r, mod) * inv(fact(r, mod) % mod, mod) % mod
end

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

puts ((pow(2, n) - 1) - comb(n, a) - comb(n, b)) % MOD