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++ (GCC 9.2.1) (judge verdict: Runtime Error)

#include <iostream>
#include <cmath>
// #include 
using namespace std;
long long rec(long long b, long long n){
    if(n < b) return n;
    else {
        return rec(b, n/b) + n%b;
     }
 }
void solve()
{
    long long n, s; cin >> n >> s;
    long long ans;
    long long b = 2;

    if(n == s) {
        ans = n + 1;
        return;
    }
    while(b <= sqrt(n)){
        long long f = rec(b, n);
        if(f == s) {
            ans = b;
            cout << b << endl;
            return;
        }
        b++;
    }
    long long p, q;
    q = 0;
    p = 1;
    while( p < sqrt(n)){
        b = (n-s)/p + 1;
        long long f = rec(b,n);
        if(f == s){
            ans = b;
            cout << ans << endl;
            return;
        }  
        p++;
    }
    cout << -1 << endl;
}

int main()
{
    solve();
    return 0;
}