CF582A GCD Table

A. GCD Table

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 xand 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.

Examples

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 

题意：给出n个数之间任意两个数的gcd,求这n个数分别是多少.

分析：我们要从这个表中得到一些原数列的信息，显然，最大的那个数肯定是原数列中的数，然后把这个数去掉，并且将这个数和之前得到的数列中的数的gcd去掉，因为是一个矩形，所以一定会出现两次gcd,重复这种操作就能得到原数列中的数.

    至于实现，最好用map、hash,如果一个一个找数会浪费很多时间.

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<cmath>
#include<map>

using namespace std;

map <int, int> m;

int a[510 * 510],n,tot,ans[510*510];

bool cmp(int x, int y)
{
    return x > y;
}

int gcd(int a, int b)
{
    if (!b)
        return a;
    return gcd(b, a % b);
}

int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n * n; i++)
    {
        scanf("%d", &a[i]);
        m[a[i]]++;
    }
    sort(a + 1, a + n * n + 1,cmp);
    for (int i = 1; i <= n * n; i++)
    {
        if (!m[a[i]])
            continue;
        m[a[i]]--;
        for (int j = 1; j <= tot; j++)
            m[gcd(a[i], a[j])] -= 2;
        ans[++tot] = a[i];
    }
    for (int i = 1; i <= tot; i++)
        printf("%d ", ans[i]);

    return 0;
}