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;
using ll = long long;
ll f(ll b, ll n)
{
    if(n < b) return n;
    else
    {
        return f(b, n/b) + n % b;
    }
}
int main()
{
    ll N, s;
    cin >> N >> s;
    if(N == s)
    {
        cout << N + 1 << endl;
        return 0;
    }

    for(ll i = 2; i * i <= N; i++)
    {
        if(f(i, N) == s)
        {
            cout << i << endl;
            return 0;
        }
    }
    for(ll i = 1; i * i <= N; i++)
    {
        ll q = s - i;
        ll b = (N - q) / i;
        if(b != 0 && f(b, N) == s)
        {            
            cout << b << endl;
            return 0;
        }
    }
    cout << -1 << endl;
}