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 OCaml (4.02.3) (judge verdict: Accepted)

let rec f acc b = function
  | 0 -> acc
  | n -> f (n mod b + acc) b (n / b)
let f = f 0

let () = Scanf.scanf "%d %d" @@ fun n s ->
  let r = int_of_float @@ sqrt @@ float_of_int n in
  if n < s
  then print_endline "-1"
  else if n = s
  then Printf.printf "%d\n" (n + 1)
  else 
    try
      Array.init r (( + ) 2)
      |> Array.to_list
      |> List.find (fun b -> f b n = s)
      |> Printf.printf "%d\n"
    with Not_found ->
      try
        Array.init ((n - s) / (r - 1)) (( + ) 1)
        |> Array.to_list
        |> List.rev_map (fun p -> (n - s) / p + 1)
        |> List.filter (( <= ) 2)
        |> List.find (fun b -> f b n = s)
        |> Printf.printf "%d\n"
      with _ -> print_endline "-1"