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>
#pragma GCC optimize("O3")
using namespace std;
const double pi = 3.141592653589793238;
const long long inf = 100000000000000000;
void pv(vector<long long> v) {
  for (long long i = 0; i < v.size(); ++i) {
    cout << v[i] << " ";
  }
  cout << endl;
}
void pa(long long a[], long long n, long long p) {
  for (long long i = p; i < n + p; ++i) {
    cout << a[i] << " ";
  }
  cout << endl;
}
long long modexp(long long x, long long y) {
  long long res = 1;
  x = x % 998244353;
  while (y > 0) {
    if (y & 1LL) res = (res * x) % 998244353;
    y = y >> 1;
    x = (x * x) % 998244353;
  }
  return res;
}
long long invmod(long long a) { return modexp(a, 998244353 - 2); }
vector<long long> edge[1000005];
void ipgraph(long long n, long long m) {
  for (long long i = 1; i <= n; i += 1) edge[i].clear();
  if (m == 0) return;
  for (long long i = 1; i <= m; i += 1) {
    long long a, b;
    cin >> a >> b;
    edge[a].push_back(b);
    edge[b].push_back(a);
  }
}
std::vector<long long> v;
unordered_map<long long, std::vector<long long>> fac;
void findFactor(long long a) {
  if (fac[a].size()) return;
  for (long long i = 1; i <= sqrt(a); i += 1) {
    if (a % i == 0) {
      fac[a].push_back(i);
      if (a / i != i) {
        fac[a].push_back(a / i);
      }
    }
  }
  sort(fac[a].begin(), fac[a].end());
  for (long long u : fac[a]) findFactor(u);
}
void dfs(long long a, long long k) {
  if (v.size() >= 100000) return;
  if (k == 0 || a == 1) {
    v.push_back(a);
    return;
  }
  for (long long u : fac[a]) {
    dfs(u, k - 1);
  }
}
void solve() {
  long long x, k;
  cin >> x >> k;
  findFactor(x);
  dfs(x, min(k, 100000LL));
  pv(v);
}
int32_t main() {
  ios ::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  for (long long i = 1; i <= t; i += 1) {
    solve();
  }
}