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<iomanip>
#include <map>
#include <algorithm>
#include <vector>
#include <string>
#include <utility>
#include <queue>
#define INF 1e9+7
#define rep(i,n) for(ll i=0;i<n;i++)
#define NO cout<<"NO"<<endl;
#define YES cout << "YES"<<endl;
#define No cout << "No"<<endl;
#define Yes cout << "Yes"<<endl;
#define all(a) a.begin(),a.end()
#define P pair<int,int>
using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;

//(int)'a'は97 (int)'A'は65 (int)'1'は49
//おまじない

const int maxn = 100000;

ll s,n;

bool f(ll b,ll n,ll S=0){
    if(S>s)return false;
    if(n<b){
        return s == S+n;
    }else{
        return f(b,n/b,S+n%b);
    }
}

int main(){
    
    cin >> n>>s;
    
    if(n==s){
        cout << n+1<<endl;
    }else{
        ll sq=2;
        
        while(sq*sq<=n){
            if(f(sq,n)&&n%sq==0){
                cout << sq<<endl;
                return 0;
            }
            sq++;
        }
        
        ll p = 1;
        while(p*p<n){
            ll b = (n-s)/p+1;
            if(b>=2&&f(b,n)){
                cout << b<<endl;
                return 0;
            }
            p++;
        }
    }
    
    cout << -1<<endl;
    
    return 0;
}