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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

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

int main(){
  long long n, s;
  cin >> n >> s;
  long long ans = -1ll;
  
  for(long long b = 2; b <= sqrt(n); b++){
    if(fnc(b,n) == s){
      ans = b;
      break;
    }
  }
  if(ans == -1ll){
    for(long long x = sqrt(n); x > 0; x--){
      if((n-s)%x == 0 && fnc((n-s)/x+1,n) == s){
        ans = (n-s)/x+1;
        break;
      }
    }
  }
  if(ans == -1ll){
    if(n == s)
      ans = n+1;
  }
    
  cout << ans << endl;
}