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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        long N = sc.nextLong();
        long S = sc.nextLong();
        System.out.println( solve(N, S) );
    }

    private static long solve(long N, long S) {
        if( N == S ) return N+1;
        if( S == 1 ) return N;
        if( N < S ) return -1;

        long rootN = (long)Math.floor(Math.sqrt(N));

        // n^2 以上で表現されるもの
        long min = Long.MAX_VALUE;
        for (long b = rootN; b > 1; b--) {
            if( f(b, N) == S ) {
                min = Math.min(b, min);
            }
        }
        if( min != Long.MAX_VALUE ) return min;

        // pb + q の形で表現されるもの
        for (long p = rootN; p >= 1; p--) {
            long b = (N - S) / p + 1;
            if( f(b, N) == S ) {
                min = Math.min(b, min);
            }
        }
        if( min != Long.MAX_VALUE ) return min;

        return -1;
    }

    private static long f(long b, long n) {
        if( n < b ) {
            return n;
        } else {
            return f(b, n/b) + (n % b);
        }
    }
}