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;
int getint() {
  unsigned int c;
  int x = 0;
  while (((c = getchar()) - '0') >= 10) {
    if (c == '-') return -getint();
    if (!~c) exit(0);
  }
  do {
    x = (x << 3) + (x << 1) + (c - '0');
  } while (((c = getchar()) - '0') < 10);
  return x;
}
int getstr(char *s) {
  int c, n = 0;
  while ((c = getchar()) <= ' ') {
    if (!~c) exit(0);
  }
  do {
    s[n++] = c;
  } while ((c = getchar()) > ' ');
  s[n] = 0;
  return n;
}
template <class T>
inline bool chmin(T &a, T b) {
  return a > b ? a = b, 1 : 0;
}
template <class T>
inline bool chmax(T &a, T b) {
  return a < b ? a = b, 1 : 0;
}
inline long long getll() {
  unsigned int c;
  long long x = 0;
  while (((c = getchar()) - '0') >= 10) {
    if (c == '-') return -getll();
    if (!~c) exit(0);
  }
  do {
    x = (x << 3) + (x << 1) + (c - '0');
  } while (((c = getchar()) - '0') < 10);
  return x;
}
inline void putint(int n) {
  int i = 16, a[i];
  if (n < 0) putchar('-');
  do {
    a[--i] = 48 + abs(n % 10);
    n /= 10;
  } while (n);
  while (i < 16) putchar(a[i++]);
}
inline void putll(long long n) {
  int i = 32, a[i];
  if (n < 0) putchar('-');
  do {
    a[--i] = 48 + abs(n % 10);
    n /= 10;
  } while (n);
  while (i < 32) putchar(a[i++]);
}
const int EXCEED = 1e5;
long long x, k;
int how_many_print_so_far = 0;
map<long long, vector<long long> > divisors;
void solve(long long current, long long depth) {
  if (current == 1) {
    printf("1 ");
    how_many_print_so_far++;
    return;
  }
  if (depth == k) {
    putll(current), putchar(' ');
    how_many_print_so_far++;
    return;
  }
  vector<long long> &divs = divisors[current];
  if (divs.size() == 0) {
    for (long long i = 1;; i++) {
      if (i * i > current) break;
      if (current % i == 0) {
        divs.push_back(i);
        if (i != current / i) {
          divs.push_back(current / i);
        }
      }
    }
    sort(divs.begin(), divs.end());
  }
  for (int i = 0; i < divs.size(); i++) {
    solve(divs[i], depth + 1);
    if (how_many_print_so_far == EXCEED) return;
  }
}
int main() {
  int i, j, tcc, tc = 1 << 28;
  for (tcc = 0; tcc < tc; tcc++) {
    x = getll(), k = getll();
    solve(x, 0);
    puts("");
  }
  return 0;
}