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 <bits/stdc++.h>
using namespace std;
typedef long long ll;

ll f(ll b,ll n){
  if (n<b) return n;
  else return f(b,n/b)+n%b;
}

ll gcd(ll a, ll b)
  {return b==0 ? a : gcd(b,a%b);}


ll ma(vector<ll> ANS){
  if (ANS.size()==0) return -1;
  else {
    sort(ANS.begin(),ANS.end());
    return ANS[0];
  }
}



int main(){
  ios_base::sync_with_stdio(false);
  cin.tie(NULL); cout.tie(NULL);
  //bool check=false;
  ll ans=-1;
  ll n,s;
  cin >> n >> s;
  ll N=n;
  if (s==1) cout<<n<<endl;
  else {
    vector<ll> divisors;
  for (int i=2; i*i<=n; i++){
    if (n%i==0) n=n/i, divisors.push_back(i);
  }
  if (n>1) divisors.push_back(n);
  vector<ll> ANS;
   

  for (ll i: divisors){
    //if (check) break;
    for (ll j=i; j<=N; j+=i){
      if (gcd(j,s)!=1 and f(j,N)==s) {
        ANS.push_back(j);
       //check=true;
        break;
      }
    }
  }
  
  cout<<ma(ANS)<<endl;
    //for (ll i: ANS) cout<<i<<" ";
  }
  



  

}