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++ (GCC 5.4.1) (judge verdict: Runtime Error)

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

int Rand(int x) {return rand() * rand() % x + 1;}

#define vi vector<int>
#define pii pair<int, int >
#define mp make_pair
#define fi first
#define se second
#define pb push_back
#define LL long long
const int INF = 0x3f3f3f3f, N = 1000005, M = 2000005, MOD = 1e9 + 7;

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

int main()
{
	ios::sync_with_stdio(false); cin.tie(0), cout.tie(0);
	LL n, m;
	while (cin >> n >> m)
	{
		if(m > n) {
			cout << -1 << endl;
			continue;
		}
		else if(n == m) {
			cout << n + 1 << endl;
			continue;
		}
		LL sn = sqrt(n), ans = -1;
		for (int i = 1; i <= sn; i++)
		{
			if (f(n, i) == m)
			{
				ans = i;
				break;
			}
		}
		sn = sqrt(n - m);
		for (int i = 1; i <= sn; i++)
			if ((n - m) % i == 0)
			{
				LL bb = (n - m) / i + 1;
				if (f(n, bb) == m)
				{
					if(ans == -1) ans = bb;
					ans = min(ans, bb);
				}
			}
		cout << ans << endl;
	}
	return 0;
}