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 ll = long long;

ll dig(ll b, ll n);

int main() {
    ll n, s;

    cin >> n >> s;

    if ((s > (n + 1) / 2 && s < n) || s > n) {
        cout << -1 << endl;
        return 0;
    }

    if (s == n) {
        cout << n + 1 << endl;
        return 0;
    }

    ll ans = 0;
    for (int i = 2; i <= n * n; i++) {
        if (dig(i, n) == s) {
            ans = i;
            break;
        }
    }

    if (ans == 0) {
        ans = n - (s - 1);
    }
    
    cout << ans << endl;
    
    return 0;
}

ll dig(ll b, ll n) {
    if (n >= b) {
        return dig(b, n / b) + n % b;
    }
    else {
        return n;
    }
}