Codeforces 894.C Marco and GCD Sequence

C. Marco and GCD Sequence

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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, ..., an. He remembered that he calculated gcd(ai, ai + 1, ..., aj) 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, ..., sm (1 ≤ si ≤ 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 < ... < sm.

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, ..., an (1 ≤ ai ≤ 106) — the sequence.

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

If there are multiple solutions, print any of them.

Examples

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(ai, ai + 1, ..., aj) for every 1 ≤ i ≤ j ≤ n.

 大致题意：给定一个序列S,由原序列任意一对(i,j)组成的区间[i,j]中的数的gcd构成，S中没有重复的数，求原序列.

分析：这种题没啥意思.i可以等于j,那么一种想法就是把所有的数放在原序列中，这样的话可能有些连续的数的gcd会出现S中没有出现过的数，解决方法也很简单，在每两个数中间插入一个s[1]即可，因为s[1]是最小的gcd,肯定是所有数的约数.据此可以判断是否有解.

#include <cstdio>
#include <cstring>
#include <iostream>
#include <stack>
#include <algorithm>

using namespace std;

int n, a[100010];
bool flag = false;

int main()
{
    scanf("%d", &n);
    scanf("%d", &a[1]);
    for (int i = 2; i <= n; i++)
    {
        scanf("%d", &a[i]);
        if (a[i] % a[1] != 0)
            flag = 1;
    }
    if (flag)
        printf("-1
");
    else
    {
        printf("%d
", n * 2);
        for (int i = 1; i <= n; i++)
            printf("%d %d ", a[i], a[1]);
    }

    return 0;
}