CodeForces 660A Co-prime Array

Co-prime Array

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 660A

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

32 7 28

Output

12 7 9 28

大意：给出一段整数序列，可以在任意位置添加任意整数，使序列每两个相邻的数都互质。

在不互质的整数间添加1。

#include<cstdio>int gcd(int a,int b){if(a%b == 0)return b;elsereturn gcd(b,a%b);}int main(){int n;while(scanf("%d",&n)!=EOF){int i,j;int a[1100],b[1100];for(i = 1 ; i  <= n ; i++){scanf("%d",&a[i]);}int ans=1,ant=0,k=0;for(i = 1 ; i < n ; i++){if(gcd(a[i],a[i+1]) != 1){ant++;}}printf("%d\n",ant);for(i = 1 ; i < n ; i++){printf("%d ",a[i]);if(gcd(a[i],a[i+1])!= 1)printf("1 ");}printf("%d\n",a[n]);}}