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 C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;

namespace ARC060_D___桁和__Digit_Sum
{
    class Program
    {
        static void Main(string[] args)
        {
            Console.WriteLine(solver());
        }
        static long digitSum(long b, long n)
        {
            long result = 0;
            while (b <= n)
            {
                result += (n % b);
                n = n / b;
            }
            result += n;
            return result;
        }
        static long solver()
        {
            var N = long.Parse(Console.ReadLine());
            var S = long.Parse(Console.ReadLine());
            var sqrtN = (long)Math.Sqrt(N);
            for (long i = 2; i < sqrtN; i++) if (digitSum(i, N) == S) return i;
            var divisorNS = new List<long>();
            for (long i = (long)Math.Sqrt(N - S); i >= 1; i--) if (((N - S) % i) == 0) divisorNS.Add((N - S) / i);
            foreach (long b in divisorNS) if (digitSum(b + 1, N) == S) return b + 1;
            return -1;
        }
    }
}