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 <bits/stdc++.h>
#define ll long long
#define fastio cin.sync_with_stdio(false); cin.tie(NULL)
using namespace std;
ll findf(ll n, ll b){
  ll re=0;
  while(n>0){
    re+=n%b;
    n/=b;
  }
  return re;
}
int main(){
  fastio;
  ll n,f;
  cin>>n>>f;
  ll ans=-1;
  if (n<=f){
  if (n==f)ans=n+1;
  if (ans==-1)
  for (ll d=2;d*d<=n;d++){
    if(findf(n,d)==f){
      ans=d;
      break;
    }
  }
  if (ans==-1){
    for (ll p=1;p*p<n;p++){
      if ((n-f)%p) continue;
      ll b=(n-f)/p+1;
      if (n/b/b==0&&f-p==n%b)
        ans=b;
    }
  }
  }  
  cout<<ans<<'\n';
  return 0;
}