Codeforces Round #447 (Div. 2) C. Marco and GCD Sequence(构造，思路)

Description

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

42 4 6 12

output

34 6 12

input

22 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.

思路

题目给出了一个严格递增的整数序列，让你构造另一个序列，满足：

新序列的任意区间求GCD都要在原序列中出现(gcd(ai, ai + 1, ..., aj) for every 1 ≤ i ≤ j ≤ n)
如果没有符合条件的序列则输出-1

首先要使新的序列能构造出来，那么原序列中最小的数肯定是其他数的因子，如果不是，那么肯定就构造不出来，构造的时候只需要把最小的数插入在原序列中间即可，这样可以保证gcd出来的结果不是自己就是最小的数

代码

#include <bits/stdc++.h>using namespace std;const int N=1000+20;int a[N];int main(){    int n;    scanf("%d",&n);    for(int i=1; i<=n; i++)        scanf("%d",&a[i]);    for(int i=1; i<=n; i++)    {        if(a[i]%a[1]!=0)        {            puts("-1");            return 0;        }    }    printf("%d\n",2*n);    for(int i=1; i<=n; i++)        printf("%d %d ",a[i],a[1]);    puts("");    return 0;}