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)

# 解説AC
include Math

def conv_bnum(b, n)
  return "error" if b < 2
  res = 0
  while n > 0
    res += n % b
    n /= b
  end
  return res
end

n = gets.to_i
s = gets.to_i
sqrt_n = sqrt(n).ceil

2.upto(sqrt_n) do |i|
  if conv_bnum(i, n) == s
    puts i
    exit
  end
end

sqrt_n.downto(1) do |i|
  j = s - i
  next if (n - j) % i != 0

  b = (n - j) / i
  next if b < 2

  str = conv_bnum(b, n)
  if s == str
    puts b
    exit
  end
end

puts -1