Codeforces Round #323 (Div. 2) C. GCD Table

C. GCD Table

The GCD table G of size n x n for an array of positive integers a of lengt H n is defined by formula

Let us remind your that the greatest common divisor (GCD) of the positive integers x and y is the greatest integer which is divisor of both xand y, it is denoted as. For example, for array a = {4, 3, 6, 2} of length 4 the GCD tablewould look as follows: /c12>

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 is positive integers, not exceeding 9. Note that the elements is given in an arbitrary order. It is guaranteed, 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 is multiple possible solutions, you is allowed to print any of them.

Examplesinput

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

Input

2
1 1 1 1

Output

One eye on the appearance of the odd number of times must be the answer, and then yy a bit, feel that there are more than two times even several times the answer, five minutes out of this problem, the result is fst (the person intentions sinister, well actually I was too weak).
Positive solution should be clearly known the largest number must be in the answer, each time the number of the introduction of the answer will be and the previous answer to some of the number of two new GCD values, the original array of the two values, to the end will leave the answer.

#include <iostream>#include<cstdio>#include<queue>#include<map>#include<algorithm>using namespacestd;#defineMAXN 620intA[MAXN *Maxn];map<int,int>MP;intMain () {intN;  while(~SCANF ("%d", &N)) { for(inti =1; I <= n * n; i++) {scanf ("%d", &A[i]); Mp[a[i]]++; } sort (A+1, A +1+ N *N); intCNT =0; intANS[MAXN];  for(inti = n * n; I >=1; i--) {            if(!mp[a[i]])Continue; Mp[a[i]]--;  for(intj =1; J <= CNT; J + +) {MP[__GCD (A[i], ans[j])]-=2; } ans[++CNT] =A[i]; }         for(inti =1; I <= CNT; i++) printf ("%d", Ans[i]); Puts""); }}

Codeforces Round #323 (Div. 2) C. GCD Table