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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>
using namespace std;

long long f(long long b, long long n) {
  if (b < 2) return -1;
  long long ans = 0;
  while (n > 0) {
    ans += n % b;
    n /= b;
  }
  return ans;
}

int main() {
  long long n, s;
  cin >> n >> s;
  if (s > n) return cout << "-1\n", 0;
  if (n == s) return cout << n + 1 << '\n', 0;

  for (long long i = 2; i <= 1e6; ++i) {
    if (f(i, n) == s) cout << i << '\n', exit(0);
  }

  long long rs = n;
  for (long long i = 1; i <= 1e6; ++i) {
    long long x = n / i;
    if (f(x, n) == s) rs = min(rs, x);
    x = n / i + 1;
    if (f(x, n) == s) rs = min(rs, x);
     x = s / i;
    if (f(x, n) == s) rs = min(rs, x);
    x = s / i + 1;
    if (f(x, n) == s) rs = min(rs, x);
     x = (n - s) / i;
    if (f(x, n) == s) rs = min(rs, x);
    x = (n - s) / i + 1;
    if (f(x, n) == s) rs = min(rs, x);
  }
  cout << rs << '\n';
  return 0;
}