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: Time Limit Exceeded)

import java.util.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);

    long n = scanner.nextLong();
    long s = scanner.nextLong();

    if (n < s) {
      System.out.println(-1);
      return;
    }

    for (int d = 39; d > 1; d--) {
      long b = lowerBound(d, n);
      while (d * Math.log(b) <= Math.log(n)) {
        if (test(b, n, s)) {
          System.out.println(b);
          return;
        }
        b++;
      }
    }

    // d = 1
    for (int a1 = (int) Math.sqrt(n - s); a1 >= 1; a1--) {
      if ((n - s) % a1 == 0) {
        long b = (n - s) / a1 + 1;
        long a0 = n - a1 * b;
        if (a0 >= 0 && a0 < b && a0 + a1 == s) {
          System.out.println(b);
          return;
        }
      }
    }

    // d = 0
    if (n == s) {
      System.out.println(n + 1);
    } else {
      System.out.println(-1);
    }
  }

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

  private static long lowerBound(int d, long n) {
    long left = 1;
    long right = n;
    double logn = Math.log(n);
    while (right - left > 1) {
      long mid = (left + right) / 2;
      double logm = (d + 1) * Math.log(mid);
      if (logm > logn) {
        right = mid;
      } else {
        left = mid;
      }
    }
    return right;
  }
}