Coprime Array

Description

You are given an array of n elements, you must make it a co-prime array in as few moves as possible.

In each move you can insert any positive integral number you want not greater than 10⁹ in any place in the array.

An array is co-prime if any two adjacent numbers of it are co-prime.

In the number theory, two integers a and b are said to be co-prime if the only positive integer that divides both of them is 1.

Input

The first line contains integer n (1 ≤ n ≤ 1000) — the number of elements in the given array.

The second line contains n integers a_i (1 ≤ a_i ≤ 10⁹) — the elements of the array a.

Output

Print integer k on the first line — the least number of elements needed to add to the array a to make it co-prime.

The second line should contain n + k integers a_j — the elements of the array a after adding k elements to it. Note that the new array should be co-prime, so any two adjacent values should be co-prime. Also the new array should be got from the original array a by addingk elements to it.

If there are multiple answers you can print any one of them.

Sample Input

Input

3
2 7 28

Output

1

2 7 9 28

给出一个序列 为了保证相邻的两个数互质（最大公约数为1） 可以再任意两个数之间插入一个数

输出最少需要插入的数的数量 和插入新数后的序列

1和任意一个数都互质 只要在不互质的两个数之间插入一个1即可

#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<algorithm>
using namespace std;
int vis[10001];
int gcd(int a,int b)
{
	 if(a%b==0)
	 return b;
	 return gcd(b,a%b);
}
int s[10000];
int main()
{
	int n;
	scanf("%d",&n);
	int sum=0;
	for(int i=0;i<n;i++)
	{
		vis[i]=0;
		scanf("%d",&s[i]);
	}
	for(int i=0;i<n-1;i++)
	{
		if(gcd(s[i],s[i+1])!=1)
		{
			vis[i]=1;
			sum++;
		}
	}
	 printf("%d\n",sum);
	for(int i=0;i<n;i++)
	{
		printf("%d",s[i]);
		if(vis[i]==1)
		printf(" 1");
		if(i!=n)
		printf(" ");
	}
	return 0;
}