Codeforces Round #565 (Div. 3)--D. Recover it!--思维+欧拉筛

D. Recover it!

Authors guessed an array aa consisting of nn integers; each integer is

not less than 22 and not greater than 2⋅1052⋅105. You don’t know the

array aa, but you know the array bb which is formed from it with the

following sequence of operations:

Firstly, let the array bb be equal to the array aa; Secondly, for each

ii from 11 to nn: if aiai is a prime number, then one integer paipai

is appended to array bb, where pp is an infinite sequence of prime

numbers (2,3,5,…2,3,5,…); otherwise (if aiai is not a prime number),

the greatest divisor of aiai which is not equal to aiai is appended to

bb; Then the obtained array of length 2n2n is shuffled and given to

you in the input. Here paipai means the aiai-th prime number. The

first prime p1=2p1=2, the second one is p2=3p2=3, and so on.

Your task is to recover any suitable array aa that forms the given

array bb. It is guaranteed that the answer exists (so the array bb is

obtained from some suitable array aa). If there are multiple answers,

you can print any.

Input

The first line of the input contains one integer nn

(1≤n≤2⋅1051≤n≤2⋅105) — the number of elements in aa.

The second line of the input contains 2n2n integers

b1,b2,…,b2nb1,b2,…,b2n (2≤bi≤27501312≤bi≤2750131), where bibi is the

ii-th element of bb. 27501312750131 is the 199999199999-th prime

number.

Output

In the only line of the output print nn integers a1,a2,…,ana1,a2,…,an

(2≤ai≤2⋅1052≤ai≤2⋅105) in any order — the array aa from which the

array bb can be obtained using the sequence of moves given in the

problem statement. If there are multiple answers, you can print any.

Examples

input

Copy

3
3 5 2 3 2 4
output

Copy

3 4 2
input

Copy

1
2750131 199999
output

Copy

199999
input

Copy

1
3 6
output

Copy

6 

题解如下

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
const int Len = 1e6;
int prime[Len * 3];
int ar[Len * 3];
int br[3 * Len];
int barrel[3 * Len];
vector<int> vec;
int n;

bool cmp(int a,int b)
{
    return a > b;
}
void Prime()
{
    for(int i = 2; i <= Len * 3; i ++)
        prime[i] = 1;
    //素数筛选法
    for(int i = 2; i * i <= Len * 3; i ++)
    {
        if(prime[i])
        {
            for(int j = i * i; j <= Len * 3; j += i)
                prime[j] = 0;
        }
    }
}

void init()
{
    Prime();

    int pos = 1;
    for(int i = 2; i <= Len * 3; i ++)
    {
        if(prime[i])
        {
            ar[pos ++] = i;
        }
    }
    //输入
    for(int i = 1; i <= 2 * n; i ++)
    {
        scanf("%d",&br[i]);
    }
    //统计各个数字出现的次数
    for(int i = 1; i <= 2 * n; i ++)
    {
        barrel[br[i]] ++;
    }
    sort(br + 1 , br + 2 * n + 1 , cmp);
}
void Solve()
{
    init();

    for(int i = 1; i <= 2 * n; i ++)
    {
        int cnt = barrel[br[i]];
        if(cnt > 0)
        {
            if(! prime[br[i]])
            {
                int mx_divisor;
                for(int j = 2; ; j ++)
                    if(br[i] % j == 0)
                    {
                        mx_divisor = br[i] / j;
                        break;
                    }

                if(barrel[mx_divisor] > 0)
                {
                    barrel[mx_divisor] --;
                    vec.push_back(br[i]);
                    barrel[br[i]] --;
                }
            }
            else
            {
                int pri = ar[br[i]];
                if(barrel[pri] > 0)
                {
                    barrel[pri] --;
                    vec.push_back(br[i]);
                    barrel[br[i]] --;
                }
            }
        }
    }

    for(auto x : vec)
        printf("%d ",x);
}

int main()
{
    //freopen("test.txt","r",stdin);
    scanf("%d",&n);
    Solve();

    return 0;
}