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: Time Limit Exceeded)

N, a, b = gets.chomp.split.map(&:to_i)
MOD = 10**9 + 7
ONE = '1'.freeze

# コンビネーションの計算
def nCk(n, k)
  (fact(k + 1, n) * inv(fact(1, n - k) % MOD)) % MOD
end

# 階乗の計算
def fact(s, e)
  (s..e).reduce(1) { |r, i| (r * i) % MOD }
end

# MOD下での逆元の計算
def inv(x)
  res = 1 # 逆元をどんどんかけるやつ
  beki = x # 逆元を計算したい値のMOD化でのべき乗
  (MOD - 2).to_s(2).reverse.chars do |digest|
    res = (beki * res) % MOD if digest == ONE
    beki = (beki * beki) % MOD
  end
  res
end

def pow(c,d)
  if d == 1
    c
  elsif d%2 == 0
    (pow(c,d/2)**2)%MOD
  else
    (c*pow(c,d/2)**2)%MOD
  end
end

all = pow(2,N)
aa = nCk(N,a)
bb = nCk(N,b)
p (all-aa-bb-1)%MOD