Coderfroces A. Co-prime Array

A. Co-prime Array

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 adding k elements to it.

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

Example

Input

3
2 7 28

Output

1
2 7 9 28
即使序列的最小公倍数相等就好了，呢么两两比较，不相等就插1；

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
ll a[2006],n;
ll gcd(ll x,ll y)
{
    return y==0?x:gcd(y,x%y);
}
int main()
{
    scanf("%d",&n);
    int ans=0;
    memset(a,0,sizeof(a));
    for(int i=1;i<=2*n-1;i+=2)
    {
        scanf("%lld",&a[i]);
    }
    for(int i=3;i<=2*n-1;i+=2)
    {
        if(gcd(a[i],a[i-2])!=1)
        {
            a[i-1]=1;
            ans++;
        }
    }
    printf("%d\n",ans);
    for(int i=1;i<=2*n-1;i++)
    {
        if(i!=1 && a[i]!=0) printf(" ");
        if(a[i]!=0) printf("%lld",a[i]);
    }
    printf("\n");
    return 0;
}