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 

 1 #include<stdio.h>
 2 #include<map>
 3 #include<algorithm>
 4 #define MAXN 20000000
 5 using namespace std;
 6 typedef long long LL;
 7 LL ys[9000];int tot=0;int tt2=0;
 8 int tail[9000];int head[9000];
 9 int aft[MAXN];LL p[MAXN];
10 LL n,k;
11 map<LL,int>bh;
12 void line(int j,int i)
13 {
14      tt2++;if(!tail[j])head[j]=tt2;p[tt2]=i;aft[tail[j]]=tt2;tail[j]=tt2;
15 }
16 void init()
17 {
18      for(LL i=1;i*i<=n;i++)
19      if(n%i==0)
20      {
21                ys[++tot]=i;
22                if(i*i!=n)
23                ys[++tot]=n/i;
24                }
25      sort(ys+1,ys+1+tot);
26      for(int i=1;i<=tot;i++)bh[ys[i]]=i;
27      for(int i=1;i<=tot;i++)
28      for(int j=1;j<=i;j++)
29      if(ys[i]%ys[j]==0)line(i,j);
30 }
31 int dfs(int now,LL dep,int need)
32 {
33     if(ys[now]==1)
34     {
35                   printf("1 ");
36                   return 1;
37                   }
38      if(dep==k)
39      {
40                int us=need;
41                for(int u=head[now];u&&need;need--,u=aft[u])
42                printf("%I64d ",ys[p[u]]);
43                return us-need;
44                }
45      int us=need;
46      for(int u=head[now];u&&need;need-=dfs(p[u],dep+1,need),u=aft[u]);
47      return us-need;
48 }
49 int main()
50 {
51     scanf("%I64d%I64d",&n,&k);if(k>100000)k=100000;
52     init();
53     if(!k)
54     {
55           printf("%I64d\n",n);
56           return 0;
57           }
58     dfs(bh[n],1,100000);
59     return 0;
60 }

Codeforces Round #256 (Div. 2) E Divisors