//25
#include<iostream>
#include<cmath>
#include<algorithm>

using namespace std;

int main(){
  for(int n;cin>>n,n;){
    long long b[90000];
    for(int i=0;i<n*(n+1)/2;i++){
      cin>>b[i];
    }
    int a=sqrt(double(b[0])*b[1]/b[n])+.5;
    cout<<a<<endl;
    int c[250];
    for(int i=0;i<n;i++){
      c[i]=b[i]/a;
    }
    sort(c,c+n);
    for(int i=0;i<n;i++){
      cout<<c[i]<<((i==n-1)?'\n':' ');
    }
  }
  return 0;
}
### Prompt

Generate a CPP solution to the following problem:

Problem H: Squid Multiplication

Squid Eiko loves mathematics. Especially she loves to think about integer. One day, Eiko found a math problem from a website.

"A sequence b ={ai + aj | i < j } is generated from a sequence a ={a0 , ... , an | ai is even if i is 0, otherwise ai is odd}. Given the sequence b , find the sequence a ."

This problem is easy for Eiko and she feels boring. So, she made a new problem by modifying this problem .

"A sequence b ={ai *aj | i < j } is generated from a sequence a ={ a0, ... , an | ai is even if  i  is 0, otherwise ai is odd}. Given the sequence b , find the sequence a ."

Your task is to solve the problem made by Eiko.



Input

Input consists of multiple datasets.
Each dataset is given by following formats.


n
b0 b1 ... bn*(n+1)/2-1


n is the number of odd integers in the sequence a. The range of n is 2 ≤ n ≤ 250. bi is separated by a space. Each bi is 1 ≤ bi ≤ 263-1. The end of the input consists of a single 0.

Output

For each dataset, you should output two lines. First line contains a0, an even number in the sequence a. The second line contains n odd elements separated by a space. The odd elements are sorted by increasing order. You can assume that the result is greater than or equal to 1 and less than or equal to 231-1.

Example

Input

3
6 10 14 15 21 35
2
30 42 35
0


Output

2
3 5 7
6
5 7

### Response

```cpp
//25
#include<iostream>
#include<cmath>
#include<algorithm>

using namespace std;

int main(){
  for(int n;cin>>n,n;){
    long long b[90000];
    for(int i=0;i<n*(n+1)/2;i++){
      cin>>b[i];
    }
    int a=sqrt(double(b[0])*b[1]/b[n])+.5;
    cout<<a<<endl;
    int c[250];
    for(int i=0;i<n;i++){
      c[i]=b[i]/a;
    }
    sort(c,c+n);
    for(int i=0;i<n;i++){
      cout<<c[i]<<((i==n-1)?'\n':' ');
    }
  }
  return 0;
}
```