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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define forx(i, a, b) for (int i = (a); i < (b); i++)
#define rep(i, n) for (int i = 0; i < (n); i++)

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);

    ll n, s;
    cin >> n >> s;
    if (s > n)
    {
        cout << -1 << endl;
        return 0;
    }
    if (s == n)
    {
        cout << n + 1 << endl;
        return 0;
    }

    for (ll b = 2; b <= (ll)floor(sqrtl(n)); b++)
    {
        ll sub = n, res = 0;
        while (sub > 0)
        {
            res += sub % b;
            sub /= b;
        }
        if (s == res)
        {
            cout << b << endl;
            return 0;
        }
    }
    vector<ll> a;
    a.push_back(1e12);

    for (ll p = 1; p <= (ll)floor(sqrtl(n)); p++)
    {
        ll b = (n - s) / p + 1;
        ll sub = n, res = 0;
        while (sub > 0)
        {
            res += sub % b;
            sub /= b;
        }
        if (s == res)
        {
            a.push_back(b);
        }
    }
    sort(a.begin(), a.end());
    if (a[0] == 1e12)
        cout << -1 << endl;
    else
        cout << a[0] << endl;

    return 0;
}