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>
#include<algorithm>
#include<vector>
#include<cmath>

using namespace std;

typedef long long ll;

vector<ll> yakusu;

void calc_yakusu(ll A) {
	
	if (A == 1) {
		yakusu.push_back(1);
		return;
	}
	ll waru = 1;
	while (waru * waru <= A) {
		if (A % waru == 0) {
			yakusu.push_back(waru);
			if(waru * waru != A)
				yakusu.push_back(A / waru);
		}
		waru++;
	}

}

int main() {
	ll n, s;
	cin >> n >> s;
	if (s == n) {
		cout << n + 1 << endl;
		return 0;
	}
	if (s > n) {
		cout << -1 << endl;
		return 0;
	}
	for (ll shinsu = 2; shinsu * shinsu <= n; shinsu++) {
		ll cpyn = n;
		ll tmp = 0;
		while (cpyn > 0)tmp += (cpyn % shinsu), cpyn /= shinsu;
		if (tmp == s) {
			cout << shinsu << endl;
			return 0;
		}
	}
	
	//n == p * shinsu + q, p + q == s
	//->q == s - p
	//n == p * shinsu + s - p
	//n - s == p * (shinsu - 1) (shinsu > sqrt(n), p, q < shinsu, p > 0, q >= 0)
	calc_yakusu(n - s);
	sort(yakusu.begin(), yakusu.end());
	
	for (int i = 0; i < yakusu.size(); i++) {
		ll cur = yakusu[i];
		ll counter = (n - s) / yakusu[i];
		if (cur + 1 > (ll)sqrt(n) && cur + 1 > counter && cur + 1 > s - counter && s - counter >= 0) {
			cout << cur + 1 << endl;
			return 0;
		}
	}
	cout << -1 << endl;
	return 0;
}