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.

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.

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 adding kelements to it.

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

#include<iostream>

#include<algorithm>

#include<cstdlib>

#include<sstream>

#include<cstring>

#include<cstdio>

#include<string>

#include<deque>

#include<cmath>

#include<queue>

#include<set>

#include<map>

using namespace std;

inline int gcd(const int &a,const int &b)

{

	return b?gcd(b,a%b):a;

}

int main (void)

{

	int n,i,j,ans,t;

	while (cin>>n)

	{

		vector<int>vec;

		map<int,int>pos;

		for (i=0; i<n; i++)

		{

			cin>>t;

			vec.push_back(t);

		}

		for (i=0; i<n-1; i++)

		{

			if(gcd(vec[i],vec[i+1])!=1)

			{

				for (j=2; j<=100000000; j++)

				{

					if(gcd(vec[i],j)==1&&gcd(vec[i+1],j)==1)//map记录每个位置是否要放入和放入的值

					{

						pos[i]=j;

						break;

					}

				}

			}

		}

		cout<<pos.size()<<endl;

		for (i=0; i<n; i++)

		{

			if(pos.find(i)!=pos.end())

				cout<<vec[i]<<" "<<pos[i];

			else

				cout<<vec[i];

			if(i==n-1)

				cout<<endl;

			else

				cout<<" ";

		}

	}

	return 0;

}