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 <string>
# include <set>
# include <map>
# include <cmath>
# include <iomanip>
# include <functional>
# include <utility>
# include <stack>
# include <queue>
# include <list>
# include <unordered_map>
using namespace std;
using LL = long long;
constexpr int MOD = 1000000000 + 7;
constexpr int INF = 2000000000;
constexpr LL LINF = 9000000000000000000;
LL f(LL b, LL n) {
	if (n < b)return n;
	else return f(b, n / b) + n%b;
}

int main() {
	LL n, s;
	cin >> n >> s;
	LL i = 2;
	for (; i <= sqrt(n); i++) {
		if (f(i, n) == s) {
			cout << i << endl;
			return 0;
		}
	}
	LL a = n - s;
	if (a < 0) {
		cout << -1 << endl;
		return 0;
	}
	else if (a == 0) {
		cout << n + 1 << endl;
		return 0;
	}
	LL ans = LINF;
	for (; i >= 1; i--) {
		if (a%i != 0)continue;
		LL b = a / i + 1;
		if (f(b, n) == s) {
			ans = min(ans, b);
		}
	}
	if (ans != LINF) {
		cout << ans << endl;
	}
	else {
		cout << -1 << endl;
	}

}