Problem:
Score : 500 points

Problem Statement

For integers b (b \geq 2) and n (n \geq 1), let the function f(b,n) be defined as follows:

f(b,n) = n, when n < b

f(b,n) = f(b,\,{\rm floor}(n / b)) + (n \ {\rm mod} \ b), when n \geq b

Here, {\rm floor}(n / b) denotes the largest integer not exceeding n / b,
and n \ {\rm mod} \ b denotes the remainder of n divided by b.

Less formally, f(b,n) is equal to the sum of the digits of n written in base b.
For example, the following hold:

f(10,\,87654)=8+7+6+5+4=30

f(100,\,87654)=8+76+54=138

You are given integers n and s.
Determine if there exists an integer b (b \geq 2) such that f(b,n)=s.
If the answer is positive, also find the smallest such b.

Constraints

1 \leq n \leq 10^{11}

1 \leq s \leq 10^{11}

n,\,s are integers.

Input

The input is given from Standard Input in the following format:

n
s

Output

If there exists an integer b (b \geq 2) such that f(b,n)=s, print the smallest such b.
If such b does not exist, print -1 instead.

Sample Input 1

87654
30

Sample Output 1

10

Sample Input 2

87654
138

Sample Output 2

100

Sample Input 3

87654
45678

Sample Output 3

-1

Sample Input 4

31415926535
1

Sample Output 4

31415926535

Sample Input 5

1
31415926535

Sample Output 5

-1

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

import Data.Maybe
import Data.List
import Data.Ratio
import Debug.Trace

main = do
  n <- readLn :: IO Integer
  s <- readLn :: IO Integer

  print $ case solve n s of
    Just x -> x
    Nothing -> -1

  where
    
    f b n
      | n < b = n
      | n >= n = (mod n b) + f b (div n b)

    maybeSeq xs = if c==[] then Nothing else Just c
      where c = catMaybes xs
      
    solve :: Integer -> Integer -> Maybe Integer
    solve n s
      | s == n = Just $ n+1
      | otherwise = fmap minimum $ maybeSeq [f1,f2]

      where
        rt = ceiling (sqrt $ fromIntegral n)
        f1 = let l = filter (\b -> f b n == s) [2..rt] in if l == [] then Nothing else Just $ minimum l
        f2 = fmap minimum $ maybeSeq $ map (\b -> solve' (div n b + (b-1)) s b) (reverse [2..rt-1])
        
        solve' x s b
           | s > x = Nothing
           | otherwise = if mod (x-s) (b-1) == 0 && div (x-s) (b-1) < (div n (b-1) - div n b) then
                           Just $ n - div n b + div (x-s) (b-1)
                         else
                           Nothing