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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <cmath>
#include <iostream>

using lli = long long;
int main(){
  lli  N,S;
  std::cin >> N >> S;

  lli ns = std::sqrt(N);

  for(int i = 2; i <= ns; i++){
    lli s = 0, n = N;
    while(n > 0){
      s += n%i;
      n /= i;
    }
    if(s == S){
      std::cout << i << std::endl;
      return 0;
    }
  }
  for(int i = ns+1; i >= 1; i--){
    lli s0 = S-i;
    if(s0 < 0) continue;
    if((N-s0)/i <= i) continue;
    if((N-s0)%i == 0 && (N-s0)/i > s0){
      std::cout << (N-s0)/i << std::endl;
      return 0;
    }
  }
  if(N == S){
    std::cout << S+1 << std::endl;
    return 0;
  }

  std::cout << "-1" << std::endl;

  return 0;
}