Educational Codeforces Round 19 A

Description

Given a positive integer n, find k integers (not necessary distinct) such that all these integers are strictly greater than 1, and their product is equal to n.

Input

The first line contains two integers n and k (2 ≤ n ≤ 100000, 1 ≤ k ≤ 20).

Output

If it's impossible to find the representation of n as a product of k numbers, print -1.

Otherwise, print k integers in any order. Their product must be equal to n. If there are multiple answers, print any of them.

Examples

input

100000 2

output

2 50000 

input

100000 20

output

-1

input

1024 5

output

2 64 2 2 2 
题意：将数字分解为k个数相乘的形式，不行输出-1
解法：先分解质因数，然后处理成k个数的形式，不行的话输出-1

#include<bits/stdc++.h>
using namespace std;
const int maxn=300000;
int n,k;
int sum;
int cmd(int n)
{
    for(int i=2; i*i<=n; i++)
    {
        if(n%i==0)
        {
            return 0;
        }
    }
    return 1;
}
int main()
{
    cin>>n>>k;
    if(k==1)
    {
        cout<<n<<endl;
        return 0;
    }
    if((n==2||cmd(n))&&k>1)
    {
        cout<<"-1"<<endl;
    }
    else
    {
        vector<int>q;
        for(int i=2; i<=n; i++)
        {
            while(n!=i)
            {
                if(n%i==0)
                {
                    q.push_back(i);
                    n=n/i;
                }
                else
                    break;
            }
        }
         q.push_back(n);
         if(q.size()<k)
         {
             cout<<"-1"<<endl;
             return 0;
         }
         for(int i=0;i<k-1;i++)
         {
             cout<<q[i]<<" ";
         }
         sum=1;
         for(int i=k-1;i<q.size();i++)
         {
             sum*=q[i];
         }
         cout<<sum<<endl;
    }
    return 0;
}