CSU-ACM2017暑假集训比赛8 - C - GCD Table - CodeForces - 582A

C - GCD Table

The GCD table G of size n × n for an array of positive integers a of length n is defined by formula

Let us remind you that the greatest common divisor (GCD) of two positive integers x and y is the greatest integer that is divisor of both x and y, it is denoted as . For example, for array a = {4, 3, 6, 2} of length 4 the GCD table will look as follows:

Given all the numbers of the GCD table G, restore array a.

Input

The first line contains number n (1 ≤ n ≤ 500) — the length of array a. The second line contains n2 space-separated numbers — the elements of the GCD table of G for array a.

All the numbers in the table are positive integers, not exceeding 109. Note that the elements are given in an arbitrary order. It is guaranteed that the set of the input data corresponds to some array a.

Output

In the single line print n positive integers — the elements of array a. If there are multiple possible solutions, you are allowed to print any of them.

Example

Input

4
2 1 2 3 4 3 2 6 1 1 2 2 1 2 3 2

Output

4 3 6 2

Input

1
42

Output

42

Input

2
1 1 1 1

Output

1 1

由最大公因数的性质可知，二维表中的最大值一定是原数组中的最大元素。由此，首先取出二维表中当前最大元素 x ，记录下来，再从表中取出次大元素 y，求两者的最大公因数 g，从表中删去两个 g，记录 y。重复上述过程，直到完成对二维表的遍历，之前记录下来的数值就是所求。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <cstring>
#include <queue>
#include <map>
using namespace std;
const int maxn = 500*500+19;

int N;
int sample[maxn], res[504];
//int test[504][504];

int main(){
#ifdef TEST
freopen("test.txt", "r", stdin);
#endif // TEST

while(cin >> N){
map<int,int> ele;
for(int i = 1; i <= N*N; i++){
scanf("%d", &sample[i]);
if(!ele.count(sample[i]))
ele[sample[i]] = 1;
else
ele[sample[i]]++;
}
sort(sample+1, sample+N*N+1);
int pos = 0;
for(int i = N*N; i > 0; i--){
if(ele[sample[i]] == 0)
continue;
ele[sample[i]]--;
for(int j = 0; j < pos; j++){
int gcd = __gcd(sample[i], res[j]);
ele[gcd] -= 2; // gcd在表中成对出现，一并去除。
}
res[pos++] = sample[i];
}
for(int i = 0; i < pos; i++)
cout << res[i] << " ";
cout << endl;
}

return 0;
}