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();
		for(long b=2;b<(long)Math.sqrt(n);b++){
			if(func(b,n)==s){
				System.out.println(b);
				return;
			}
		}
		//n=p*b+q
		//s=p+q
		//n-s=p*(b-1)
		for(long i=1;i<=n-s;i++){
			if((n-s)%i==0 && n-s/i+1>Math.sqrt(n))System.out.println(n-s/i+1);
		}
		System.out.println(-1);
	}
	
	static long func(long b,long n){
		if(n<b)return n;
		else {
			long retval=0;
			while(n>=b){
				retval+=n%b;
				n=n/b;
			}
			retval+=n%b;
			return retval;
		}
	}
}