Problem: You are given two integers n and k. Find k-th smallest divisor of n, or report that it doesn't exist.

Divisor of n is any such natural number, that n can be divided by it without remainder.

Input

The first line contains two integers n and k (1 ≤ n ≤ 1015, 1 ≤ k ≤ 109).

Output

If n has less than k divisors, output -1.

Otherwise, output the k-th smallest divisor of n.

Examples

Input

4 2


Output

2


Input

5 3


Output

-1


Input

12 5


Output

6

Note

In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.

In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn't exist, so the answer is -1.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long n, k, cnt, s[10000001];
inline long long read() {
  long long w = 1, q = 0;
  char ch = ' ';
  while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-') w = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') q = q * 10 + ch - '0', ch = getchar();
  return w * q;
}
inline void write(long long x) {
  if (x < 0) {
    x = ~(x - 1);
    putchar('-');
  }
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
signed main() {
  n = read();
  k = read();
  if (n == 1) {
    if (k == 1)
      write(1);
    else
      write(-1);
    return 0;
  }
  for (long long i(1); i <= sqrt(n); i++) {
    if (n % i == 0) {
      s[++cnt] = i;
      if (i != n / i) s[++cnt] = n / i;
    }
  }
  sort(s + 1, s + cnt + 1);
  if (cnt < k)
    write(-1);
  else
    write(s[k]);
  return 0;
}