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>
inline int F() {
  register int aa, bb, ch;
  while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-')
    ;
  ch == '-' ? aa = bb = 0 : (aa = ch - '0', bb = 1);
  while (ch = getchar(), ch >= '0' && ch <= '9') aa = aa * 10 + ch - '0';
  return bb ? aa : -aa;
}
inline long long Fll() {
  register long long aa;
  int bb, ch;
  while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-')
    ;
  ch == '-' ? aa = bb = 0 : (aa = ch - '0', bb = 1);
  while (ch = getchar(), ch >= '0' && ch <= '9') aa = aa * 10 + ch - '0';
  return bb ? aa : -aa;
}
long long p[31622776];
int cnt = 0;
int main() {
  long long n = Fll(), k = Fll(), lim = sqrt(n);
  for (int i = 1; i <= lim; ++i)
    if (n % i == 0) p[++cnt] = i;
  if (cnt >= k)
    printf("%I64d\n", p[k]);
  else {
    for (int i = cnt; i; --i)
      if (n / p[i] != p[i]) p[++cnt] = n / p[i];
    if (cnt >= k)
      printf("%I64d\n", p[k]);
    else
      puts("-1");
  }
}