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: Accepted)

#include <bits/stdc++.h> 

using namespace std;
using ll=long long;
using ull=unsigned long long;
using pii=pair<int,int>;

#define INF LONG_MAX
#define MOD 1000000007
#define rng(a) a.begin(),a.end()
#define rrng(a) a.end(),a.begin()

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

int main(){
  ios::sync_with_stdio(false);
  cin.tie(0);

  //b進数のnの桁和がs
  ll n,s;
  cin>>n>>s;
  if(n==s){
    cout<<n+1<<endl;
    return 0;
  }
  if(n<s){
    cout<<-1<<endl;
    return 0;
  }
  ll ans=LONG_MAX;
  for(ll i=2;i*i<=n;i++){
    if(s==f(i,n))ans=min(ans,i);
  }
  //n=bp+q
  //s=p+q
  //b=s/p (b>sqrt(n))
  for(ll i=1;i*i<n;i++){
    ll b=(n-s)/i+1;
    if(s==n/b+n%b)ans=min(ans,b);
  }

  if(ans==LONG_MAX)cout<<-1<<endl;
  else cout<<ans<<endl;
  return 0;
}