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 INF = 1 << 29;
const long long LINF = 1LL << 60;
const double inf = 1e15;
long long mod = 1e9 + 7;
char READ_DATA;
int SIGNAL_INPUT;
template <typename Type>
inline Type ru(Type &v) {
  SIGNAL_INPUT = 1;
  while ((READ_DATA = getchar()) < '0' || READ_DATA > '9')
    if (READ_DATA == '-')
      SIGNAL_INPUT = -1;
    else if (READ_DATA == EOF)
      return EOF;
  v = READ_DATA - '0';
  while ((READ_DATA = getchar()) >= '0' && READ_DATA <= '9')
    v = v * 10 + READ_DATA - '0';
  v *= SIGNAL_INPUT;
  return v;
}
inline long long modru(long long &v) {
  long long p = 0;
  SIGNAL_INPUT = 1;
  while ((READ_DATA = getchar()) < '0' || READ_DATA > '9')
    if (READ_DATA == '-')
      SIGNAL_INPUT = -1;
    else if (READ_DATA == EOF)
      return EOF;
  p = v = READ_DATA - '0';
  while ((READ_DATA = getchar()) >= '0' && READ_DATA <= '9') {
    v = (v * 10 + READ_DATA - '0') % mod;
    p = (p * 10 + READ_DATA - '0') % (mod - 1);
  }
  v *= SIGNAL_INPUT;
  return p;
}
template <typename A, typename B>
inline char ru(A &x, B &y) {
  if (ru(x) == EOF) return EOF;
  ru(y);
  return 2;
}
template <typename A, typename B, typename C>
inline char ru(A &x, B &y, C &z) {
  if (ru(x) == EOF) return EOF;
  ru(y);
  ru(z);
  return 3;
}
template <typename A, typename B, typename C, typename D>
inline char ru(A &x, B &y, C &z, D &w) {
  if (ru(x) == EOF) return EOF;
  ru(y);
  ru(z);
  ru(w);
  return 4;
}
struct Edge {
  int u, v, next;
  long long w, cap, flow;
  Edge(int _u = 0, int _v = 0, int nxt = -1, long long _w = 1,
       long long _cap = 0) {
    u = _u;
    v = _v;
    w = _w;
    cap = _cap;
    flow = 0;
    next = nxt;
  }
  int operator<(const Edge &b) const { return w < b.w; }
};
const int maxn = 1e6 + 2, N = 1e5, S = 300;
double eps = 1e-7, pi = acos(-1.0);
unsigned long long seed = 131, smod = (1LL << 32) - 267;
long long x, k;
int cnt;
map<long long, int> f;
int tot;
vector<long long> d[maxn];
void divide(long long n) {
  vector<long long> tmp;
  int m = sqrt(n);
  for (int i = 1; i <= m; ++i)
    if (n % i == 0) {
      tmp.push_back(i);
      if (n / i != i) tmp.push_back(n / i);
    }
  sort(tmp.begin(), tmp.end());
  m = tmp.size();
  for (register int i = 0; i < m; ++i) f[tmp[i]] = i;
  for (register int i = 0; i < m; ++i)
    for (register int j = 0; j <= i; ++j)
      if (tmp[i] % tmp[j] == 0) {
        d[i].push_back(tmp[j]);
      }
}
void dfs(long long u, long long k) {
  if (u == 1 || k == 0) {
    cout << u << " ";
    ++cnt;
    return;
  }
  int id = f[u], m = d[id].size();
  for (register int i = 0; i < m; ++i) {
    dfs(d[id][i], k - 1);
    if (cnt == N) return;
  }
}
int main() {
  ru(x, k);
  divide(x);
  dfs(x, k);
}