title: K Th Prime Factor Given Number
signature: kThPrimeFactorGivenNumber :: Integer -> Integer -> Double
  kThPrimeFactorGivenNumber n k = 
test_code: import Text.Printf
  import Control.Exception (catch, SomeException)
  import Data.Typeable
  import qualified Data.List as List
  import qualified Data.Map as Map
  import qualified Data.Set as Set
  
  -- SOLUTION CODE
  -- ============================================
  
  PLACEHOLDER_CODE_BODY
  
  -- TESTING CODE 
  -- ============================================
  validateSolution :: Double -> Double -> Bool
  validateSolution actual expected = do
      (abs (actual-expected)) < 1e-06
  
  driver :: Integer -> Integer -> Double -> String
  driver n k expected = do
      if (validateSolution (PLACEHOLDER_FN_NAME n k) expected)
          then "PASSED"
      else "FAILED"
  
  main = do
      printf "TEST-0...%s\n" (driver (94) (0) (-1.0) )
  
      printf "TEST-1...%s\n" (driver (99) (1) (3.0) )
  
      printf "TEST-2...%s\n" (driver (64) (3) (2.0) )
  
      printf "TEST-3...%s\n" (driver (27) (3) (3.0) )
  
      printf "TEST-4...%s\n" (driver (24) (4) (3.0) )
  
      printf "TEST-5...%s\n" (driver (84) (6) (-1.0) )
  
      printf "TEST-6...%s\n" (driver (69) (98) (-1.0) )
  
      printf "TEST-7...%s\n" (driver (69) (39) (-1.0) )
  
      printf "TEST-8...%s\n" (driver (22) (60) (-1.0) )
  
      printf "TEST-9...%s\n" (driver (39) (57) (-1.0) )
  
source_py: def k_th_prime_factor_given_number(n, k):
      while ((n % 2) == 0):
          k = (k - 1)
          n = (n / 2)
          if (k == 0):
              return 2
      i = 3
      while (i <= math.sqrt(n)):
          while ((n % i) == 0):
              if (k == 1):
                  return i
              k = (k - 1)
              n = (n / i)
          i = (i + 2)
      if ((n > 2) and (k == 1)):
          return n
      return (- 1)
source_cpp: int k_th_prime_factor_given_number(int n, int k) {
    while ( n % 2 == 0 ) {
      k --;
      n = n / 2;
      if ( k == 0 ) return 2;
    }
    for ( int i = 3;
    i <= sqrt ( n );
    i = i + 2 ) {
      while ( n % i == 0 ) {
        if ( k == 1 ) return i;
        k --;
        n = n / i;
      }
    }
    if ( n > 2 && k == 1 ) return n;
    return - 1;
  }