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.io.BufferedReader;
import java.io.InputStreamReader;

class Main {
	public static void main(String[] args) throws java.lang.Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		long n = Long.parseLong(br.readLine());
		long s = Long.parseLong(br.readLine());
		Solver sol = new Solver();
		System.out.println(sol.solve(n, s));
	}

}

class Solver {

	long solve(long n, long s) {
		long temp;
		if (s > n)
			return -1;
		if (s == n)
			return n + 1;
		for (long b = 2; b * b <= n; b++) {
			if (findSum(b, n) == s) {
				return b;
			}
		}

		for (long i = 1; i * i < n; i++) {
			temp = ((n - s) / i) + 1;
			if (findSum(temp, n) == s) {
				return temp;
			}
		}
		return -1L;
	}

	long findSum(long b, long n) {
		long temp = 0L;
		while (n >= b) {
			temp += n % b;
			n /= b;
		}
		return temp + n;

	}

}