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

# 繰り返し二乗法(x.pow(n, mod)でもOK)
def pow(x, n, mod = MOD)
  ans = 1
  # 下位ビットから繰り返し計算を行う
  while n > 0 do
    ans = ans * x % mod if n.odd? # 下位ビットが1かどうか
    x = x * x % mod
    n >>= 1 # 右ビットシフト
  end
  return ans
end

# (mod p)下での逆元
# フェルマーの小定理より a^(p-1) = 1 (mod p) (aは整数, pは素数, aとpは互いに素)
# a * a^(p-2) = 1 なので, aの逆元が a^(p-2) となる
def inv(a, mod = MOD)
  pow(a, mod - 2, mod)
end

# 階乗
# 再利用できるようグローバル変数化
$fact = [1] # 0! = 1
def fact(n, mod = MOD)
  # 計算済ならそれを返す
  return $fact[n] unless $fact[n].nil?

  # 階乗を求めながら$factへ格納
  f = $fact[-1]
  ($fact.size).upto(n) do |i|
    f = (f * i) % mod
    $fact << f
  end
  return $fact[n]
end

# 順列
def perm(n, r, mod = MOD)
  # n*(n-1)*(n-2)*...*(n-r+1)
  return n.downto(n - r + 1).reduce(1) {|ans, x| (ans * x) % mod }
end

# 組み合わせ
def comb(n, r, mod = MOD)
  return 0 unless 0 <= r && r <= n
  return 1 if r == 0 || r == n
  # nCk = n!/(n-k)! * 1/k!
  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