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;
long long DIV[1 << 20], aux[1 << 20];
int cnt, aux_cnt, idx = 0;
long long out[1 << 20];
int P[1 << 20], prime_cnt;
int not_prime[1 << 20];
int dp_prime[1 << 20];
vector<long long> divs[1 << 20];
void sieve() {
  int mx = 1 << 20;
  for (int i = 2; i < mx; i++)
    if (not_prime[i] == 0) {
      P[prime_cnt++] = i;
      if (i * 1LL * i >= mx) continue;
      for (int j = i * i; j < mx; j += i) not_prime[j] = 1;
    }
}
bool is_prime(long long n) {
  int idx = lower_bound(DIV, DIV + cnt, n) - DIV;
  if (dp_prime[idx] != -1) return dp_prime[idx];
  for (int i = 0; i < (prime_cnt); i++) {
    if (P[i] * 1LL * P[i] > n) break;
    if (n % P[i] == 0) return dp_prime[idx] = 0;
  }
  return dp_prime[idx] = n > 1;
}
void get_DIVs(long long n, long long *DIV, int &cnt) {
  int idx = lower_bound(::DIV, ::DIV + ::cnt, n) - ::DIV;
  if (::cnt && !divs[idx].empty()) {
    copy(divs[idx].begin(), divs[idx].end(), DIV);
    cnt = divs[idx].size();
    return;
  }
  cnt = 0;
  for (long long i = 1; i * i <= n; i++)
    if (n % i == 0) {
      DIV[cnt++] = i;
      if (i * i != n) DIV[cnt++] = n / i;
    }
  sort(DIV, DIV + cnt);
  divs[idx] = vector<long long>(DIV, DIV + cnt);
}
void work(long long n, long long k, int &mx) {
  if (k == 0) {
    if (mx) out[idx++] = n, mx--;
  } else if (n == 1) {
    if (mx) {
      out[idx++] = 1;
      mx--;
    }
  } else if (is_prime(n)) {
    for (int i = 0; i < k && mx; i++, mx--) out[idx++] = 1;
    work(n, 0, mx);
  } else {
    get_DIVs(n, aux, aux_cnt);
    vector<long long> DIVs(aux, aux + aux_cnt);
    for (int i = 0; i < ((int)DIVs.size()) && mx; i++) work(DIVs[i], k - 1, mx);
  }
}
int main() {
  sieve();
  memset(dp_prime, -1, sizeof dp_prime);
  long long n, k;
  cin >> n >> k;
  int mx = 100 * 1000;
  get_DIVs(n, DIV, cnt);
  work(n, k, mx);
  for (int i = 0; i < (idx); i++) printf("%lld%c", out[i], " \n"[i == idx - 1]);
  return 0;
}