題目鏈接：https://cn.vjudge.net/problem/CodeForces-894C

In a dream Marco met an elderly man with a pair of black glasses. The man told him the key to immortality and then disappeared with the wind of time.

When he woke up, he only remembered that the key was a sequence of positive integers of some length n, but forgot the exact sequence. Let the elements of the sequence be a

Note that even if a number is put into the set S twice or more, it only appears once in the set.

Now Marco gives you the set S and asks you to help him figure out the initial sequence. If there are many solutions, print any of them. It is also possible that there are no sequences that produce the set S

Input

The first line contains a single integer m (1?≤?m?≤?1000) — the size of the set S.

The second line contains m integers s₁,?s₂,?...,?s_m (1?≤?s_i?≤?10⁶) — the elements of the set S. It‘s guaranteed that the elements of the set are given in strictly increasing order, that means s

Output

If there is no solution, print a single line containing -1.

Otherwise, in the first line print a single integer n denoting the length of the sequence, n should not exceed 4000.

In the second line print n integers a₁,?a₂,?...,?a_n (1?≤?a_i?≤?10⁶) — the sequence.

We can show that if a solution exists, then there is a solution with n not exceeding 4000 and a_i not exceeding 10⁶.

If there are multiple solutions, print any of them.

Example

4
2 4 6 12

3
4 6 12

2
2 3

-1

Note

In the first example 2?=?gcd(4,?6), the other elements from the set appear in the sequence, and we can show that there are no values different from 2, 4, 6 and 12 among gcd(a_i,?a_i?+?1,?...,?a_j) for every 1?≤?i?≤?j?≤?n.

題意：

有一個數組a[1~n]，對他們所有的1<=i<=j<=n求 gcd( a[i] ~ a[j] )，得到集合S；

該集合S滿足：元素不重復、集合內元素滿足嚴格單增；

現在給你一個S，讓你求出a；

題解：

gcd( a[1] ~ a[n] )顯然是所有gcd( a[i] ~ a[j] )裏最小的且滿足 gcd( a[1] ~ a[n] ) | ?gcd( a[i] ~ a[j] )，所以在集合S中S[1]應該滿足 S[1] | S[i] ;

然後另外一個性質是gcd(num) = num，所以所有的a[i]都應該出現在S裏；

我們當然不能像題目裏樣例那樣求a[1~n]，這樣有點難，考慮另外的方法；

考慮讓每個gcd(a[i])=S[i]，然後讓gcd(a[i]~a[j])=S[1]（i<j），怎麽操作呢，在S[2]~S[m]之間都插入S[1]即可。

AC代碼：

#include <bits/stdc++.h>
using namespace std;

int m,S[1005];
int main()
{
    cin>>m;
    for(int i=1;i<=m;i++) scanf("%d",&S[i]);

    bool ok=1;
    for(int i=2;i<=m;i++)
    {
        if(S[i]%S[1]!=0)
        {
            ok=0;
            break;
        }
    }
    if(!ok)
    {
        printf("-1\n");
        return 0;
    }

    printf("%d\n", m + ( (m-1==0)?(0):(m-2) ) );
    printf("%d ",S[1]);
    for(int i=2;i<=m;i++)
    {
        if(i!=2) printf(" %d ",S[1]);
        printf("%d",S[i]);
    }
    cout<<endl;
}

codeforces 894C - Marco and GCD Sequence - [有關gcd數學題]