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>
using namespace std;

using int64 = long long;

int64 f(int64 b, int64 n)
{
    if (n < b) return n;
    else return n % b + f(b, n / b);
}

int main()
{
    int64 n, s;
    cin >> n >> s;

    int64 ans = -1;
    for (int64 b = 2; b <= min(1000000LL, n); b++)
    {
        if (f(b, n) == s)
        {
            ans = b;
            break;
        }
    }

    // xy
    for (int64 x = 1; x <= min(1000000LL, s); x++)
    {
        int64 y = s - x;
        // x*b+y==nを満たすbがあるかどうか
        if (n - y < 0 or (n - y) % x != 0)
        {
            continue;
        }
        int64 b = (n - y) / x;
        if (b > x and b > y and (ans < 0 or b < ans))
        {
            ans = b;
        }
    }

    cout << ans << endl;

    return 0;
}