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.split.map(&:to_i)


MOD = (10**9) + 7 # い つ も の
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


s = 0
c1 = 1
c2 = 1

for i in 1..n
    c1 *= (n - i + 1)
    c2 *= i
    c1 %= MOD

    ncr = c1 *inv(c2) % MOD
#        p [c1, c2, ncr]
    if i != a && i != b
        s += ncr
        s %= 1000000007
    end
end
p s