Codeforces Round #256 (Div. 2) E Divisors

E. Divisors

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 X_i, for integer i (i ≥ 0): X₀ = [X] ([X] is a sequence consisting of a single number X), X_i = f(X_i - 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 X_k. 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 X_k in a single line, separated by a space. If the number of elements exceeds 10⁵, then print only the first 10⁵ elements.

Sample test(s)

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 

#include<stdio.h>
#include<map>
#include<algorithm>
#define MAXN 20000000
using namespace std;
typedef long long LL;
LL ys[9000];int tot=0;int tt2=0;
int tail[9000];int head[9000];
int aft[MAXN];LL p[MAXN];
LL n,k;
map<LL,int>bh;
void line(int j,int i)
{
     tt2++;if(!tail[j])head[j]=tt2;p[tt2]=i;aft[tail[j]]=tt2;tail[j]=tt2;
}
void init()
{
     for(LL i=1;i*i<=n;i++)
     if(n%i==0)
     {
               ys[++tot]=i;
               if(i*i!=n)
               ys[++tot]=n/i;
               }
     sort(ys+1,ys+1+tot);
     for(int i=1;i<=tot;i++)bh[ys[i]]=i;
     for(int i=1;i<=tot;i++)
     for(int j=1;j<=i;j++)
     if(ys[i]%ys[j]==0)line(i,j);
}
int dfs(int now,LL dep,int need)
{
    if(ys[now]==1)
    {
                  printf("1 ");
                  return 1;
                  }
     if(dep==k)
     {
               int us=need;
               for(int u=head[now];u&&need;need--,u=aft[u])
               printf("%I64d ",ys[p[u]]);
               return us-need;
               }
     int us=need;
     for(int u=head[now];u&&need;need-=dfs(p[u],dep+1,need),u=aft[u]);
     return us-need;
}
int main()
{
    scanf("%I64d%I64d",&n,&k);if(k>100000)k=100000;
    init();
    if(!k)
    {
          printf("%I64d
",n);
          return 0;
          }
    dfs(bh[n],1,100000);
    return 0;
}