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

using ll = long long;

const ll INF = 2e15;

ll func(ll b, ll n) {
	if (b == 1) {
		return -1;
	}
	ll res = 0;
	while (n) {
		res += n % b;
		n /= b;
	}
	return res;
}

int main() {
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);

	ll N, S; cin >> N >> S;
	ll V = ll(sqrt(N));

	ll ans = INF;

	auto check = [&](ll b) -> void {
		if (func(b, N) == S) {
			ans = min(ans, b);
		}
	};

	for (ll b = 2; b <= V; b++) {
		check(b);
	}

	check(N + 1);

	ll D = N-S;
	if (D > 0) {
		for (ll x = 1; x <= V; x++) {
			if (D % x) continue;
			check(x + 1);
			check(D / x + 1);
		}
	}

	if (ans == INF) ans = -1;
	cout << ans << '\n';

	return 0;
}