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 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <cmath>
#define inf 1e13+1
using namespace std;
long long int f(long long int b,long long int n) 
{
    if (n<b) 
        return n;
    else 
        return f(b,floor(n/b))+(n%b);
}
long long int n,s,b,p,ans=inf;
int main(void) 
{
    cin>>n>>s;
    if (n==s) 
    {
        cout<<n+1;
        return 0;
    }
    long long int sqrtn = sqrt(n);
    for (b=2;b<=sqrtn;b++) 
        if (s==f(b,n)) 
        {
            cout<<b;
            return 0;
        }
    for (p=1;p<sqrtn;p++) 
    {
        if ((s-n)%p!=0) 
            continue;
        b=(n-s)/p+1;
        if (b<2) 
            continue;
        if (s==(f(b,n))) 
            ans=min(ans,b);
    }
    if (ans<inf) cout<<ans;
    else cout<<-1;
    return 0;
}