codeforces-894C Marco and GCD Sequence

C. Marco and GCD Sequence

time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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 rem embered 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.

题意：一个序列是求出原来的任意两个数(i<j)的gcd(a[i],a[j]),如果gcd不在序列中就把它放进去。求原来的序列，会有多个答案。

思路：任意两个数的gcd，最小的肯定会放在第一个，如果第一个不能被整除的话，结果就是-1。如果是的话，就把第一个数放入序列中

#include<stdio.h>#include<string>#include<iostream>#include<math.h>#define ll long long#define maxn 1003using namespace std;int a[maxn];int main(){    int n;    cin>>n;    cin>>a[0];    for(int i=1;i<n;i++)    {        cin>>a[i];        if(a[i]%a[0]!=0)        {            cout<<-1<<endl;return 0;        }    }    cout<<2*n-1<<endl;    for(int i=0;i<n-1;i++)    {        cout<<a[i]<<" "<<a[0]<<" ";    }    cout<<a[n-1]<<endl;}