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 Haskell (GHC 7.10.3) (judge verdict: Accepted)

import Control.Applicative

main :: IO ()
main = do
    [n, a, b] <- (map read . words) <$> getLine
    print $ mod ((pow 2 n) - 1 - comb n a - comb n b) prime

pow :: Int -> Int -> Int
pow x n
    | n == 0 = 1
    | mod n 2 == 0 = pow (mod (x * x) prime) (div n 2)
    | otherwise = mod (x *  pow x (n - 1)) prime

fact :: Int -> Int
fact 1 = 1
fact n = n * fact (n - 1)

comb :: Int -> Int -> Int
comb n r = do
     (denominator n r) * (pow (numerator n r) (prime - 2))

modMul :: Int -> Int -> Int
modMul x y = mod (x * y) prime

denominator :: Int -> Int -> Int
denominator n r = foldl modMul 1 [n - r + 1..n]

numerator :: Int -> Int -> Int 
numerator n r = foldl modMul 1 [2..r]

prime :: Int
prime = 10 ^ 9 + 7