Problem:
Score : 600 points

Problem Statement

Given is a positive integer N.

We will choose an integer K between 2 and N (inclusive), then we will repeat the operation below until N becomes less than K.

Operation: if K divides N, replace N with N/K; otherwise, replace N with N-K.

In how many choices of K will N become 1 in the end?

Constraints

2 \leq N \leq 10^{12}

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of choices of K in which N becomes 1 in the end.

Sample Input 1

6

Sample Output 1

3

There are three choices of K in which N becomes 1 in the end: 2, 5, and 6.

In each of these choices, N will change as follows:

When K=2: 6 \to 3 \to 1

When K=5: 6 \to 1

When K=6: 6 \to 1

Sample Input 2

3141

Sample Output 2

13

Sample Input 3

314159265358

Sample Output 3

9

Solution in Haskell (GHC 7.10.3) (judge verdict: Wrong Answer)

import qualified Data.Vector.Unboxed as VU
import qualified Data.Vector.Unboxed.Mutable as VUM
import qualified Data.ByteString.Char8 as BS
import Control.Monad
import Data.List
import Data.Char


main :: IO ()
main = do
    n <- readLn
    let xx = 2 + (length $ divisors (n-1))
    let ys = divisors n
    let yy = length $ filter ((==) 1) $ zipWith (\x y -> y `mod` x) ys $ map (f n) ys
    print $ xx + yy


f :: Int -> Int -> Int
f n k = if n `mod` k == 0 then f (n`div`k) k else n


divisors :: Int -> [Int]
divisors n = go [] 2 
    where go :: [Int] -> Int -> [Int]
          go xs count 
              | count > (floor $ sqrt (fromIntegral n)) = xs
              | n `mod` count == 0                      = if n /= count^2 then go ((n`div`count):count:xs) (count+1) else count:xs
              | otherwise                               = go xs (count+1)