Codeforces 894C

题目链接：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 a₁, a₂, ..., a_n. He remembered that he calculated gcd(a_i, a_i + 1, ..., a_j) for every 1 ≤ i ≤ j ≤ n and put it into a set S. gcd here means the greatest common divisor.

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, in this case print -1.

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 s₁ < s₂ < ... < s_m.

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

Input

4
2 4 6 12

Output

3
4 6 12

Input

2
2 3

Output

-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.

题意：

有一个数组a[1~n]，对他们所有的1<=i<=j<=n求 gcd( a[i] ~ a[j] )，得到集合S；

该集合S满足：元素不重复、集合内元素满足严格单增；

现在给你一个S，让你求出a；

题解：

gcd( a[1] ~ a[n] )显然是所有gcd( a[i] ~ a[j] )里最小的且满足 gcd( a[1] ~ a[n] ) | ∀gcd( a[i] ~ a[j] )，所以在集合S中S[1]应该满足 S[1] | S[i] ;

然后另外一个性质是gcd(num) = num，所以所有的a[i]都应该出现在S里；

我们当然不能像题目里样例那样求a[1~n]，这样有点难，考虑另外的方法；

考虑让每个gcd(a[i])=S[i]，然后让gcd(a[i]~a[j])=S[1]（i<j），怎么操作呢，在S[2]~S[m]之间都插入S[1]即可。

AC代码：

#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
");
        return 0;
    }

    printf("%d
", 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;
}