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 <iostream>
using namespace std;
long long f(long long b,long long n)
{
	if (b==1)
	return -1;
	if (!n)
	return 0;
	return f(b,n/b)+n%b;
}
int main()
{
	long long n,s,mini=(1LL<<60);
	cin >> n >> s;
	for (long long i=2;i*i<=n;i++)
	{
		if (f(i,n)==s)
		mini=min(mini,i);
	}
	if (n==s)
	mini=min(mini,n+1);
	long long d=n-s;
	for (long long i=1;i*i<=d;i++)
	{
		if (d%i==0)
		{
			if (f(i+1,n)==s)
			mini=min(mini,i+1);
			if (f(d/i+1,n)==s)
			mini=min(mini,d/i+1);
		}
	}
	if (mini==(1LL<<60))
	mini=-1;
	cout << mini;
}