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 long long _abs(const long long x) { return x >= 0 ? x : -x; }
inline long long _max(const long long x, const long long y) {
  return x >= y ? x : y;
}
inline long long _min(const long long x, const long long y) {
  return x <= y ? x : y;
}
inline long long _gcd(const long long x, const long long y) {
  return y ? _gcd(y, x % y) : x;
}
inline long long _lcm(const long long x, const long long y) {
  return x * y / _gcd(x, y);
}
inline long long read() {
  long long fh = 1, x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') fh = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') {
    x = (x << 3) + (x << 1) + ch - '0';
    ch = getchar();
  }
  return fh * x;
}
inline void write(const long long x) {
  long long top = 0, cp = _abs(x);
  char stk[25];
  if (x == 0) {
    putchar('0');
    return;
  }
  while (cp) stk[++top] = (cp % 10) + 48, cp /= 10;
  if (x < 0) stk[++top] = '-';
  while (top) putchar(stk[top--]);
}
long long n, k, a[30000007], cnt;
signed main() {
  long long n = read(), k = read();
  long long t = std::sqrt(n);
  for (long long i = 1; i <= t; i++)
    if (n % i == 0) a[++cnt] = i;
  for (long long i = t - (t * t == n); i >= 1; i--)
    if (n % i == 0) a[++cnt] = n / i;
  if (cnt < k)
    write(-1);
  else
    write(a[k]);
  return 0;
}