#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;

#define LL long long
LL of[309],jf[90009], p[309];

int main(){
	int n;
	while(scanf("%d",&n),n){
		int ol=0,jl=0;
		for(int i=0; i<(n+1)*n/2;i++){
			LL x;
			scanf("%lld",&x);
			if(x%2==0)
				of[ol++] = x;
			else
				jf[jl++] = x;
		}
		sort(of,of+ol);
		sort(jf,jf+jl);
		
		LL a0 = int(sqrt((1.0*of[0]/jf[0]*of[1]))+0.001);
		printf("%lld\n%lld",a0,of[0]/a0);
		for(int i=1;i<n;i++)
			printf(" %lld",of[i]/a0);
		printf("\n");
	}
	return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;

#define LL long long
LL of[309],jf[90009], p[309];

int main(){
	int n;
	while(scanf("%d",&n),n){
		int ol=0,jl=0;
		for(int i=0; i<(n+1)*n/2;i++){
			LL x;
			scanf("%lld",&x);
			if(x%2==0)
				of[ol++] = x;
			else
				jf[jl++] = x;
		}
		sort(of,of+ol);
		sort(jf,jf+jl);
		
		LL a0 = int(sqrt((1.0*of[0]/jf[0]*of[1]))+0.001);
		printf("%lld\n%lld",a0,of[0]/a0);
		for(int i=1;i<n;i++)
			printf(" %lld",of[i]/a0);
		printf("\n");
	}
	return 0;
}
```