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 powmod(x, n, m)
  if n == 0
    1
  elsif n % 2 == 0
    powmod(x * x % m, n / 2, m)
  else
    powmod(x * x % m, n / 2, m) * x % m
  end
end

def inverse(x)
  powmod(x, MOD - 2, MOD)
end

def fact(n)
  (1..n).inject(1) {|a, b| (a * b) % MOD }
end

def falling_fact(n, k)
  (n-k+1..n).inject(1) {|a, b| (a * b) % MOD }
end

def binom(n, k)
  falling_fact(n, k) * inverse(fact(k))
end

n, a, b = gets.split(" ").map(&:to_i)
puts (powmod(2, n, MOD) - 1 - binom(n, a) - binom(n, b)) % MOD