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[];int tot=;int tt2=;
 int tail[];int head[];
 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=;i*i<=n;i++)
      if(n%i==)
      {
                ys[++tot]=i;
                if(i*i!=n)
                ys[++tot]=n/i;
                }
      sort(ys+,ys++tot);
      for(int i=;i<=tot;i++)bh[ys[i]]=i;
      for(int i=;i<=tot;i++)
      for(int j=;j<=i;j++)
      if(ys[i]%ys[j]==)line(i,j);
 }
 int dfs(int now,LL dep,int need)
 {
     if(ys[now]==)
     {
                   printf("1 ");
                   return ;
                   }
      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+,need),u=aft[u]);
      return us-need;
 }
 int main()
 {
     scanf("%I64d%I64d",&n,&k);if(k>)k=;
     init();
     if(!k)
     {
           printf("%I64d\n",n);
           return ;
           }
     dfs(bh[n],,);
     return ;
 }