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

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

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.

#include <bits/stdc++.h>using namespace std;int a[10001];int gcd(int a, int b){return b == 0 ? a : gcd(b, a % b);}int main(){int n;cin >> n;for(int i = 1; i <= n; ++i){scanf("%d", &a[i]);}int cur = a[1];for(int i = 2; i <= n; ++i){cur = gcd(cur, a[i]);}if(cur != a[1]){cout << -1 << endl;return 0;}cout << (n * 2) << endl;for(int i = 1; i <= n; ++i){printf("%d %d ", a[1], a[i]);}}/*题意：1000个数，每个数不超过1e6，让你构造一个不超过4000的数列，使得对于新数列，任意区间内的数求GCD都出现在这1000个数中。思路：首先排除无解的情况：如果所有数的GCD不等于最小的数，那么一定无解。如果存在解，如何构造呢，如果仅仅把原序列输出的话，可能会错，只需在原数列的相邻两个数中插入原序列最小的数即可。这样可以保证gcd不是自己就是最小的数。*/