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: Wrong Answer)

####### 二項係数 C(n,k) mod pの計算 #######

MOD=10**9+7

## a**b mod p by 繰り返し二乗法 ##
def pow(a,b)
  return 1 if b==0
  return a*pow(a,b-1)%MOD if b%2==1
  return pow(a**2%MOD,b/2)
end

## n! mod p ##
def fact(n)
  ret=1
  n.times do |i|
    ret=ret*(i+1)%MOD
  end
  ret
end

## a**(-1) mod p ##
def inv(a)
  pow(a,MOD-2)%MOD
end

## C(n,k) mod p ##
def com(n,k)
  (fact(n)*inv(fact(k)))%MOD*inv(fact(n-k))%MOD
end

##########################################


N,A,B=gets.split.map(&:to_i)
a=1
b=1

for num in 0...A do
  a=(a*(N-num))%MOD
end

for num in 0...A do
  a=(a*(A-num))%MOD
end


for num in 0...B do
  b=(b*(N-num))%MOD
end

for num in 0...A do
  b=(b*(B-num))%MOD
end


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