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.ArrayList;
import java.util.Collections;
import java.util.List;
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(ans(n,s));
	}

	static long ans(long n, long s) {
		if(n<s) {
			return -1;
		} else {
			List<Long> divisor = new ArrayList<Long>();
			divisor.add(1L);
			for(long i=2;i<=Math.sqrt(n-s);i++) {
				if(digitSum(n,i)==s) {
					return i;
				}
				if((n-s)%i == 0) {
					divisor.add(i);
				}
			}
			Collections.reverse(divisor);
			for (Long l : divisor) {
				if(digitSum(n,(n-s)/l+1)==s) {
					return (n-s)/l+1;
				}
			}
			return -1;
		}
	}

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

}