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: Wrong Answer)

N = gets.to_i 
S = gets.to_i
R = Math.sqrt(N).floor

def largest
  N == S ? N + 1 : nil 
end

def smaller
  ans = nil 
  (2..R).each do |b| 
    tot = 0
    n = N
    while n > 0 
      tot += (n % b)
      n /= b 
    end
    if tot == S 
      ans = b 
      break
    end
  end
  ans 
end

def larger
  ans = nil
  R.downto(1) do |x| # x * b + y = N, x + y = S
    if (N - S) % x == 0
      b = (N - S) / x + 1
      y = S - x
      if 0 <= y && y < b
        ans = b
        break 
      end
    end
  end
  ans
end

puts largest || smaller || larger || -1