Codeforces Round #629 (Div. 3) D.Carousel

Codeforces Round #629 (Div. 3) D.Carousel

The round carousel consists of n figures of animals. Figures are numbered from 1 to n in order of the carousel moving. Thus, after the n-th figure the figure with the number 1 follows. Each figure has its own type — the type of the animal corresponding to this figure (the horse, the tiger and so on). The type of animal of the i-th figure equals ti.

The example of the carousel for n=9 and t=[5,5,1,15,1,5,5,1,1].
You want to color each figure in one of the colors. You think that it’s boring if the carousel contains two different figures (with the distinct types of animals) going one right after another and colored in the same color.

Your task is to color the figures in such a way that the number of distinct colors used is the minimum possible and there are no figures of the different types going one right after another and colored in the same color. If you use exactly k distinct colors, then the colors of figures should be denoted with integers from 1 to k.

Input

The input contains one or more test cases.

The first line contains one integer q (1≤q≤104) — the number of test cases in the test. Then q test cases follow. One test case is given on two lines.

The first line of the test case contains one integer n (3≤n≤2⋅105) — the number of figures in the carousel. Figures are numbered from 1 to n in order of carousel moving. Assume that after the n-th figure the figure 1 goes.

The second line of the test case contains n integers t1,t2,…,tn (1≤ti≤2e5), where ti is the type of the animal of the i-th figure.

The sum of n over all test cases does not exceed 2e5.

Output

Print q answers, for each test case print two lines.

In the first line print one integer k — the minimum possible number of distinct colors of figures.

In the second line print n integers c1,c2,…,cn (1≤ci≤k), where ci is the color of the i-th figure. If there are several answers, you can print any.

Example

input

4
5
1 2 1 2 2
6
1 2 2 1 2 2
5
1 2 1 2 3
3
10 10 10

output

2
1 2 1 2 2
2
2 1 2 1 2 1
3
2 3 2 3 1
1
1 1 1 

比赛时有点懵，其实就差一点就A了，/(ㄒoㄒ)/~~
首先颜色最多只有3种~
下面讨论几种情况：
1.当所有元素相同时，很简单~
2.当 $n$ 为偶数时，只需两种颜色，以 $[1,2,1,2\cdots ]$循环即可
3.当 $n$ 为奇数时：如果所有相邻的种类都不同，前 $n-1$ 个数以 $[1,2,1,2\cdots ]$，最后输出一个3即可；若有两个相邻的元素相等，那么我们只需把这两个元素标为统一颜色，其余部分以 $[1,2,1,2\cdots ]$循环上色即可，因为我们相当于把两个元素合并为一个，总元素数就变为 $n-1$ 个，此时为偶数，对应第2点
AC代码如下：

#include 
using namespace std;
typedef long long ll;
const int N=2e5+5;
int main(){
    int q,n;
    cin>>q;
    while(q--){
        int cnt=1;
        cin>>n;
        int a[n+1],ans[n+1]={0};
        for(int i=0;i<n;i++) cin>>a[i];
        for(int i=0;i<n-1;i++){
            if(a[i]==a[i+1]) cnt++;
        }
        if(cnt==n){
            cout<<1<<endl;
            for(int i=0;i<n;i++) cout<<1<<" ";
            puts("");
            continue;
        }
        else if(n%2==0){
            cout<<2<<endl;
            for(int i=0;i<n;i++) cout<<i%2+1<<" ";
            puts("");
            continue;
        }
        else{
            int pos=-1;
            for(int i=0;i<n;i++){
                if(a[i]==a[(i+1)%n]) {pos=i;break;}
            }
            if(pos!=-1){
                for(int i=0,j=pos+1;j<n;j++,i^=1) ans[j]=i+1;
                for(int i=0,j=pos;j>=0;j--,i^=1) ans[j]=i+1;
                cout<<2<<endl;
                for(int i=0;i<n;i++) cout<<ans[i]<<" ";
                puts("");
                continue;
            }
            cout<<3<<endl;
            for(int i=0;i<n-1;i++) cout<<i%2+1<<" ";
            cout<<3<<endl;
            continue;
        }
    }
}