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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

#define INF (INT32_MAX)
#define MOD (1000000000+7)
#define MAX 50

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

int main() {
	long long n, s;

	cin >> n >> s;

	long long ans = n + 1;
	if( n != s ) {
		double rn = sqrt( n );
		int flag = 0;
		for( long long b = 2; b <= rn; b++ ) {
			if( f( b, n ) == s ) {
				ans = b;
				flag = 1;
				break;
			}
		}
		if( flag == 0 ) {
			for( int p = 1; p < rn; p++ ) {
				long long b = (n - s) / p + 1;
				if( f( b, n ) == s ) {
					ans = b;
					break;
				}
			}
		}
		if( f( ans, n ) != s ) {
			ans = -1;
		}
	}

	cout << ans << endl;

	return 0;
}