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;
template <class T>
void debug(T a, T b) {
  for (; a != b; ++a) cerr << *a << ' ';
  cerr << endl;
}
template <class T>
void chmin(T& a, const T& b) {
  if (a > b) a = b;
}
template <class T>
void chmax(T& a, const T& b) {
  if (a < b) a = b;
}
namespace std {
template <class S, class T>
ostream& operator<<(ostream& out, const pair<S, T>& a) {
  out << '(' << a.first << ',' << a.second << ')';
  return out;
}
}  // namespace std
long long int x;
long long int k;
int MX = 100000;
long long int res[100005];
int cnt;
vector<long long int> divs;
long long int prime[30];
int cnt2;
void push(long long int a) {
  res[cnt++] = a;
  if (cnt >= MX) {
    for (int i = 0; i < (cnt); ++i) cout << res[i] << ' ';
    cout << endl;
    exit(0);
  }
}
void dfs(long long int x, long long int k) {
  if (k == 0) {
    push(x);
    return;
  }
  if (x == 1) {
    push(x);
    return;
  }
  if (binary_search(prime, prime + cnt2, x)) {
    for (int i = 0; i < k; ++i) push(1);
    push(x);
    return;
  }
  for (int i = 0; i < (divs.size()); ++i)
    if (x % divs[i] == 0) {
      dfs(divs[i], k - 1);
    }
}
int main() {
  cin >> x >> k;
  if (x == 1) {
    puts("1");
    return 0;
  }
  for (long long int i = 2; i * i <= x; ++i)
    if (x % i == 0) {
      divs.push_back(i);
      if (i * i != x) divs.push_back(x / i);
    }
  divs.push_back(x);
  divs.push_back(1);
  sort((divs).begin(), (divs).end());
  for (int i = 0; i < (divs.size()); ++i) {
    int flag = 0;
    for (int j = (1); j < (divs.size()); ++j)
      if (i != j && divs[i] % divs[j] == 0) flag = 1;
    if (!flag) prime[cnt2++] = divs[i];
  }
  dfs(x, k);
  for (int i = 0; i < (cnt); ++i) cout << res[i] << ' ';
  cout << endl;
  return 0;
}