Divisors CodeForces - 448E

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本文介绍了一种使用递归方法求解特定数学序列Xk的算法，该序列涉及整数的因数分解和序列操作。通过输入整数X和k，算法能够计算并输出Xk序列的前10^5个元素。

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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

#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cstdlib>
using namespace std;
typedef long long ll;
ll a[100005],cnt,p,k,f;
void dfs(ll n,ll num)
{
    if(cnt >= 100000)
    {
        f = 1;
        return ;
    }
    if(k == num||n == 1)
    {
        cnt++;
        printf("%lld ",n);
        return ;
    }
    for(ll i = 0;i<p;i++)
    {
        if(a[i] > n) break;
        if(n % a[i] == 0)
        {
            dfs(a[i],num+1);
            if(f) return ;
        }
    }
    return ;
}
int main()
{
    ll n,i,j;
    scanf("%lld%lld",&n,&k);
    p = 0;
    for(i = 1;i*i<=n;i++)
    {
        if(n%i == 0)
        {
            a[p++] = i;
            if(i != n/i)
                a[p++] = n/i;
        }
    }
    sort(a,a+p);
    f = 0;
    cnt = 0;
    dfs(n,0);
    return 0;
}