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: Runtime Error)

import java.util.Scanner;

/**
 * https://abc044.contest.atcoder.jp/tasks/arc060_b
 */
public class Main {

	public static void main(String[] args) {
		
		Scanner sc = new Scanner(System.in);
		long n = sc.nextLong();
		long s = sc.nextLong();
		sc.close();
		
		long ans = -1;
		if(s==n){
			ans = n+1;
		}
		long sqrt = (long)Math.sqrt(n);
		if(ans==-1){
			for(long b=2; b<=sqrt; b++){
				if(s==compute(n,b)){
					ans = b;
					break;
				}
			}
		}	
		if(ans==-1){
			for(long p=sqrt; p>=1; p--){
				long b = (n-s+p)/p;
				if(s==compute(n,b)){
					ans = b;
					break;
				}
			}	
		}
		System.out.println(ans);
	}

	private static long compute(long n, long b) {
		long sum = 0;
		while(n>0){
			sum += n%b;
			n = n/b;
		}
		return sum;
	}

}