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)

Scanf.scanf "%d %d" (fun n s ->
    let rec f b n = if n < b then n else f b (n/b) + (n mod b) in

    let step3 n =
        if n = s then n + 1 else -1
    in

    let step2 n =
        let ns = n - s in
        let rec loop a =
            if a = 0 then step3 n else
                if ns mod a = 0 && s - a <= ns / a then ns / a + 1 else
                    loop (a - 1)
        in
        loop (min s (int_of_float (floor (sqrt (float ns) +. 0.5))))
    in

    let rec loop i =
        if i * i > n then step2 n else
            if f i n = s then i else
                loop (i + 1)
    in
    loop 2 |> Printf.printf "%d\n"
)