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: Wrong Answer)

#include <iostream>
#include <climits>
using namespace std;

long long digitsum(long long b, long long n) {
    long long result = 0;
    while (n > 0) {
        result += n % b;
        n /= b;
    }
    return result;
}

int main() {
    long long n, s, result = LLONG_MAX;
    cin >> n >> s;
    for (long long b = 2; b * b <= n; b++) {
        if (digitsum(b, n) == s) {
            result = b;
            break;
        }
    }
    for (long long b = 2; b * b <= n - s; b++) {
        if ((n - s) % b == 0) {
            if (digitsum((n - s) / b, n) == s) {
                result = min(result, (n - s) / b);
            }
            if (digitsum(b, n) == s) {
                result = min(result, b);
            }
        }
    }
    if (n - s > 1 && digitsum(n - s, n) == s) {
        result = min(result, n - s);
    }
    if (s == 1) {
        result = min(result, n);
    }
    if (result == LLONG_MAX) {
        cout << -1;
    } else {
        cout << result;
    }
    return 0;
}