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;
#define ll long long
#define l long
#define pb push_back
#define mp make_pair
#define mt make_tuple

bool check(ll n,ll b,ll s)
{
    ll ans = 0;
    while(n)
    {
        ans += n%b;
        n/=b;
    }
    
    return (ans == s) ;
    
}


int main() {
	// your code goes here
	//ifstream cin("input.txt");
	//ofstream cout("output.txt");
	ios_base::sync_with_stdio(false); 
    cin.tie(NULL); 
	
	ll n,s; cin>>n>>s;
	if(s>n){ cout<<-1; return 0; }
	//cout<<check(n,10,s)<<'\n';
	for(ll b=2;b<=sqrt(n);b++)
	{
	    if( check(n,b,s) ){ cout<<b; return 0; }
	}
	if(n==s && n<=9){ cout<<10; return 0; }
	for(ll i=2;i<=sqrt(n-s);i++)
	{
	    if((n-s)%i==0)
	    {
	        if( check(n,i,s) ){ cout<<i; return 0; }
	        if( check(n,(n-s)/i,s) ){ cout<<(n-s)/i; return 0; }
	    }
	}
	if( n!=1 && check(n,n,s) ){ cout<<n; return 0; }
	cout<<-1;
	
	return 0;
}