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=;
 int n,k;
 int sum;
 int cmd(int n)
 {
     for(int i=; i*i<=n; i++)
     {
         if(n%i==)
         {
             return ;
         }
     }
     return ;
 }
 int main()
 {
     cin>>n>>k;
     if(k==)
     {
         cout<<n<<endl;
         return ;
     }
     if((n==||cmd(n))&&k>)
     {
         cout<<"-1"<<endl;
     }
     else
     {
         vector<int>q;
         for(int i=; i<=n; i++)
         {
             while(n!=i)
             {
                 if(n%i==)
                 {
                     q.push_back(i);
                     n=n/i;
                 }
                 else
                     break;
             }
         }
          q.push_back(n);
          if(q.size()<k)
          {
              cout<<"-1"<<endl;
              return ;
          }
          for(int i=;i<k-;i++)
          {
              cout<<q[i]<<" ";
          }
          sum=;
          for(int i=k-;i<q.size();i++)
          {
              sum*=q[i];
          }
          cout<<sum<<endl;
     }
     return ;
 }