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>
#define endl "\n"
using namespace std;

vector< long long > ans;
long long n, s, d;

long long sum( long long x, long long base )
{
	long long ret = 0;
	while( x )
	{
		ret += x%base;
		x /= base;
	}
	return ret;	
}
int main()
{
	ios_base::sync_with_stdio( 0 );
	cin.tie( 0 );
	cin>>n>>s;
	long long k = 500000;
	for( long long a = 2; a <= k; a++ )
	{
		if( sum( n, a ) == s )
		{
			cout<<a<<endl;
			return 0;
		}
	}
	if( n == s )
	{
		cout<<n+1;
		return 0;	
	}
	for( long long j = 0; j < k; j++ )
	{
		if( j == s )
		{
//			cout<<j<<endl;
			if( n == s )ans.push_back( j+1 );
			break;	
		}
		d = s - j;
		if( ( n - j ) % d == 0 && ( ( n - j) / d ) > j && ( ( n - j) / d ) > 1 )
		{
			ans.push_back( ( ( n - j ) / d ) );	
		}
	}
	sort( ans.begin(), ans.end() );
	if( ans.size() )cout<<ans[0]<<endl;
	else cout<<"-1"<<endl;
	return 0;
}