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 Ruby (2.3.3) (judge verdict: Accepted)

N = gets.to_i
S = gets.to_i

def calc(b, n)
    n < b ? n : n%b + calc(b,n/b)
end

def calc_two_digit(b, n)
    n%b + (n/b)
end

2.upto(989898) do |b|
    if calc(b, N) == S
        p b; exit
    end
end

best = nil
1.upto(989898) do |s_high|
    # s EQ n%b + s_high
    # n EQ n%b + s_high*b
    # s EQ n-s_high*b + s_high
    # s_high*b EQ n - s + s_high
    # b EQ (n - s + s_high)/s_high
    b = (N - S + s_high)/s_high
    if b > 1 && calc(b, N) == S
        best = b if !best || b < best
    end
end

p best ? best : N == S ? S+1 : S == 1 ? N : -1