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;
void snum(long long &x) { scanf("%lld", &x); }
void snum(long long &x, long long &y) { scanf("%lld %lld", &x, &y); }
void snum(long long &x, long long &y, long long &z) {
  scanf("%lld %lld %lld", &x, &y, &z);
}
struct Factor {
  long long base, exp;
};
vector<Factor> v;
long long factorize(long long x) {
  long long resp = 1;
  long long i = 2;
  while (i * i <= x) {
    long long exp = 0;
    while (x % i == 0) {
      exp++;
      x /= i;
    }
    resp *= (exp + 1);
    if (exp > 0) {
      Factor F;
      F.base = i, F.exp = exp;
      v.push_back(F);
    }
    i++;
  }
  if (x > 1) {
    resp *= 2;
    Factor a;
    a.base = x;
    a.exp = 1;
    v.push_back(a);
  }
  return resp;
}
long long maximo;
vector<long long> lista;
void f(long long pos, long long acum) {
  if (pos == v.size()) return;
  f(pos + 1, acum);
  long long it = 0;
  while (acum <= maximo / v[pos].base && it < v[pos].exp) {
    acum *= v[pos].base;
    lista.push_back(acum);
    f(pos + 1, acum);
    it++;
  }
}
int main() {
  long long n, k;
  snum(n, k);
  maximo = n;
  if (k > factorize(n)) {
    printf("-1\n");
    return 0;
  }
  lista.push_back(1);
  f(0, 1);
  sort(lista.begin(), lista.end());
  printf("%lld\n", lista[k - 1]);
  return 0;
}