Problem: Bizon the Champion isn't just friendly, he also is a rigorous coder.

Let's define function f(a), where a is a sequence of integers. Function f(a) returns the following sequence: first all divisors of a1 go in the increasing order, then all divisors of a2 go in the increasing order, and so on till the last element of sequence a. For example, f([2, 9, 1]) = [1, 2, 1, 3, 9, 1].

Let's determine the sequence Xi, for integer i (i ≥ 0): X0 = [X] ([X] is a sequence consisting of a single number X), Xi = f(Xi - 1) (i > 0). For example, at X = 6 we get X0 = [6], X1 = [1, 2, 3, 6], X2 = [1, 1, 2, 1, 3, 1, 2, 3, 6].

Given the numbers X and k, find the sequence Xk. As the answer can be rather large, find only the first 105 elements of this sequence.

Input

A single line contains two space-separated integers — X (1 ≤ X ≤ 1012) and k (0 ≤ k ≤ 1018).

Output

Print the elements of the sequence Xk in a single line, separated by a space. If the number of elements exceeds 105, then print only the first 105 elements.

Examples

Input

6 1


Output

1 2 3 6 


Input

4 2


Output

1 1 2 1 2 4 


Input

10 3


Output

1 1 1 2 1 1 5 1 1 2 1 5 1 2 5 10 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 7;
const int INF = 1e9 + 7;
const long long int LINF = 1e18;
const long long int MOD = 1e9 + 7;
using namespace std;
int C, T;
vector<long long int> DV[5000];
map<long long int, int> M;
inline int div(long long int n) {
  if (M.count(n)) {
    return M[n];
  }
  vector<long long int> tmp;
  for (long long int i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      DV[T].push_back(i);
      if (i * i != n) {
        tmp.push_back(n / i);
      }
    }
  }
  while (!tmp.empty()) {
    DV[T].push_back(tmp.back());
    tmp.pop_back();
  }
  tmp.clear();
  M[n] = T++;
  return T - 1;
}
void go(long long int n, int k) {
  if (n == 1ll || !k) {
    printf("%I64d ", n);
    C++;
    if (C == 100000) {
      exit(0);
    }
  } else {
    int D = div(n);
    for (auto i : DV[D]) {
      go(i, k - 1);
    }
  }
}
int main() {
  long long int n;
  int k;
  cin >> n >> k;
  go(n, k);
  return 0;
}