Codeforces Round #256 (Div. 2) E. Divisors 因子+dfs

E. Divisors

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 

 题意：给你一个数，分解k次；

　　　每次将每个数分解成它的质因数，个数>=1e5不输出；

思路：枚举质因数，dfs求解，详见代码；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
const int N=2e5+10,M=4e6+10,inf=1e9+10;
ll x,k,flag;
set<ll>s;
set<ll>::iterator it;
ll p[N],len;
void dfs(ll x,ll step)
{
    if(step>k)return;
    if(flag>=100000)return;
    if(step==k||x==1)
    {
        if(flag>=100000)return;
        printf("%lld ",x);
        flag++;
        return;
    }
    for(ll i=0;i<len;i++)
    {
        ll v=p[i];
        if(v>x)break;
        //cout<<v<<"cccc"<<x<<endl;
        if(x%v==0)
        dfs(v,step+1);
    }
}
int main()
{
    scanf("%lld%lld",&x,&k);
    len=0;
    for(ll i=1;i*i<=x;i++)
    {
        if(i*i==x)
        p[len++]=i;
        else if(x%i==0)
        p[len++]=i,p[len++]=x/i;
    }
    sort(p,p+len);
    flag=0;
    dfs(x,0);
    return 0;
}