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: Wrong Answer)

#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
using ll = long long;

ll n,s;
ll res = -1;
void solve()
{
    ll sum = 0;
    ll tmp = 0;
    for(ll b = 2;b<=(ll)(sqrt(n));b++){
        sum = 0;
        tmp = n;
        while(tmp!=0){
            sum+=tmp%b;
            tmp/=b;
        }
        if(sum==s){
            res = b;
            return;
        }
    }
    if(n == s){
        res = n+1;
        return;
    }
    if(n < s){
        return ;
    }
    else{
        //find all factors of (n-s)
        ll b;
        for(ll f = 1;f<=(ll)(sqrt(n-s));f++){
            if((n-s)%f)continue;
            //f is b-1
            //f is q, (n-s)//f is (b-1)
            b = (n-s)/f+1;
            if(n%b+n/b==s && b>(ll)(sqrt(n))){
                res = b;
                return;
            }
        }
    }
}


int main()
{
    cin>>n>>s;
    solve();
    cout<<res<<endl;
    return 0;
}