Problem: 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 t_i.

<image> 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 ≤ 10^4) — 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 ⋅ 10^5) — 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 t_1, t_2, ..., t_n (1 ≤ t_i ≤ 2 ⋅ 10^5), where t_i is the type of the animal of the i-th figure.

The sum of n over all test cases does not exceed 2⋅10^5.

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 c_1, c_2, ..., c_n (1 ≤ c_i ≤ k), where c_i 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 

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
import java.text.*;

public class Main {

	// solution credits : Ayush Kumar

	void solve(int TC) {
		int n=ni();
		int[]a=new int[n];
		boolean same=true;
		for(int i=0;i<n;i++) {
			a[i]=ni();
			if(i>0&&a[i]!=a[i-1])same=false;
		}
		if(same) {
			pn("1");
			for(int i=0;i<a.length;i++)p("1 ");
			return;
		}
		
		if(n%2==0) {
			pn("2");
			for(int i=0;i<a.length;i++)p(i%2+1+" ");
			return;
		}
		boolean sa=false;
		for(int i=0;i<a.length;i++)if(a[i]==a[(i+1)%a.length])sa=true;
		if(sa) {
			int[]ans=new int[n];
			ans[0]=1;
			for(int i=1;i<n;i++) {
				if(a[i]==a[i-1]&&sa) {
					ans[i]=ans[i-1];
					sa=false;
				} else {
					ans[i]=ans[i-1]==1?2:1;
				}
			}
			pn("2");
			for(int v:ans)p(v+" ");
			return;
		}
		
		pn("3");
		for(int i=0;i<a.length-1;i++)p(i%2+1+" ");
		p("3");
		
	}

	// solution ends
	long mod = (long) 1e9 + 7, IINF = (long) 1e17, SZ = (long) 1e5;
	final int MAX = (int) 1e6 + 1, INF = (int) 2e9, root = 3;
	DecimalFormat df = new DecimalFormat("0.000000000000");
	double PI = 3.1415926535897932384626433832792884197169399375105820974944, eps = 1e-8;
	static boolean multipleTC = true, memory = false;
	FastReader in;
	PrintWriter out;

	void run() throws Exception {
		in = new FastReader();
		out = new PrintWriter(System.out);
		int T = (multipleTC) ? ni() : 1;
		for (int i = 1; i <= T; i++) {
			solve(i);
			pn("");
		}
		out.flush();
		out.close();
	}

	public static void main(String[] args) throws Exception {
		if (memory)
			new Thread(null, new Runnable() {
				public void run() {
					try {
						new Main().run();
					} catch (Exception e) {
						e.printStackTrace();
					}
				}
			}, "1", 1 << 28).start();
		else
			new Main().run();
	}

	long gcd(long a, long b) {
		return (b == 0) ? a : gcd(b, a % b);
	}

	int gcd(int a, int b) {
		return (b == 0) ? a : gcd(b, a % b);
	}

	int lcm(int[] a) {
		int rv = a[0];
		for (int i = 1; i < a.length; i++) {
			rv = (rv * a[i]) / gcd(rv, a[i]);
		}
		return rv;
	}

	int bit(long n) {
		return (n == 0) ? 0 : (1 + bit(n & (n - 1)));
	}

	void p(Object o) {
		out.print(o);
	}

	void pn(Object o) {
		out.println(o);
	}

	void pni(Object o) {
		out.println(o);
		out.flush();
	}

	String n() {
		return in.next();
	}

	String nln() {
		return in.nextLine();
	}

	int ni() {
		return Integer.parseInt(in.next());
	}

	long nl() {
		return Long.parseLong(in.next());
	}

	double nd() {
		return Double.parseDouble(in.next());
	}

	public long powMOD(long x, long n) {
		if (n == 0 || n == 1)
			return n == 0 ? 1 : x;
		long res = powMOD(x, n / 2);
		return n % 2 == 0 ? mul2(res, res) : mul3(x, res, res);
	}

	public long pow(int x, int n) {
		if (n == 0 || n == 1)
			return n == 0 ? 1 : x;
		long res = powMOD(x, n / 2);
		return n % 2 == 0 ? res * res : x * res * res;
	}

	public long add(long a, long b) {
		return ((a % mod) + (b % mod)) % mod;
	}

	public long mul2(long a, long b) {
		return ((a % mod) * (b % mod)) % mod;
	}

	public long mul3(long a, long b, long c) {
		long i = 1;
		i = mul2(a, i);
		i = mul2(b, i);
		i = mul2(c, i);
		return i;
	}

	<T> void par(T[] a) {
		for (T val : a)
			p(val + " ");
	}

	void pmat(int[][] a) {
		for (int i = 0; i < a.length; i++) {
			for (int j = 0; j < a[i].length; j++) {
				p(a[i][j] + " ");
			}
			pn("");
		}
	}

	void narr(int[] a) {
		for (int i = 0; i < a.length; i++)
			a[i] = ni();
	}

	long arrSum(int[] a) {
		long rv = 0;
		for (int val : a)
			rv += val;
		return rv;
	}

	int[] ps(int[] a) {
		int[] ps = new int[a.length];
		ps[0] = a[0];
		for (int i = 1; i < a.length; i++)
			ps[i] = ps[i - 1] + a[i];
		return ps;
	}

	int[] toArr(ArrayList<Integer> al) {
		int[] rv = new int[al.size()];
		for (int i = 0; i < rv.length; i++)
			rv[i] = al.get(i);
		return rv;
	}

	ArrayList<Integer> toAl(int[] a) {
		ArrayList<Integer> al = new ArrayList<>();
		for (int val : a)
			al.add(val);
		return al;
	}

	void swap(int i, int j, int[] a) {
		a[i] ^= a[j];
		a[j] ^= a[i];
		a[i] ^= a[j];
	}

	long findSqrt(long x) {
		long l = 0, h = x / 2, ans = -1;
		while (l <= h) {
			long m = l + ((h - l) / 2);
			if (m * m <= x) {
				ans = m;
				l = m + 1;
			} else {
				h = m - 1;
			}
		}
		return ans;
	}

	class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		public FastReader(String s) throws Exception {
			br = new BufferedReader(new FileReader(s));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}
}