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 PYTHON3):

Solution:
import sys
import os
from io import BytesIO, IOBase
from collections import Counter, deque
import math
'''import time
import collections
import itertools
import timeit
import random
from bisect import bisect_left as bl
from bisect import bisect_right as br '''
#########################
# imgur.com/Pkt7iIf.png #
#########################

# returns the list of prime numbers less than or equal to n:
'''def sieve(n):
    if n < 2: return list()
    prime = [True for _ in range(n + 1)]
    p = 3
    while p * p <= n:
        if prime[p]:
            for i in range(p * 2, n + 1, p):
                prime[i] = False
        p += 2
    r = [2]
    for p in range(3, n + 1, 2):
        if prime[p]:
            r.append(p)
    return r'''

# returns all the divisors of a number n(takes an additional parameter start):
'''def divs(n, start=1):
    divisors = []
    for i in range(start, int(math.sqrt(n) + 1)):
        if n % i == 0:
            if n / i == i:
                divisors.append(i)
            else:
                divisors.extend([i, n // i])
    return divisors'''

# returns the number of factors of a given number if a primes list is given:
'''def divn(n, primes):
    divs_number = 1
    for i in primes:
        if n == 1:
            return divs_number
        t = 1
        while n % i == 0:
            t += 1
            n //= i
        divs_number *= t
    return divs_number'''

# returns the leftmost and rightmost positions of x in a given list d(if x isnot present then returns (-1,-1)):
'''def flin(d, x, default=-1):
    left = right = -1
    for i in range(len(d)):
        if d[i] == x:
            if left == -1: left = i
            right = i
    if left == -1:
        return (default, default)
    else:
        return (left, right)'''

# returns (b**p)%m 
'''def modpow(b,p,m):
    res=1
    b=b%m
    while(p):
        if p&1:
            res=(res*b)%m
        b=(b*b)%m
        p>>=1
    return res'''

# if m is a prime this can be used to determine (1//a)%m or modular multiplicative inverse of a number a
'''def mod_inv(a,m):
    return modpow(a,m-2,m)'''

# returns the ncr%m for (if m is a prime number) for very large n and r
'''def ncr(n,r,m):
    res=1
    if r==0:
        return 1
    if n-r<r:
        r=n-r
    p,k=1,1
    while(r):
        res=((res%m)*(((n%m)*mod_inv(r,m))%m))%m
        n-=1
        r-=1
    return res'''

# returns ncr%m (if m is a prime number and there should be a list fact which stores the factorial values upto n):
'''def ncrlis(n,r,m,fact):
    return (fact[n]*(mod_inv(fact[r],m)*mod_inv(fact[n-r],m))%m)%m'''


def ceil(n, k): return n // k + (n % k != 0)
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return list(map(int, input().split()))
def lcm(a, b): return abs(a * b) // math.gcd(a, b)
def prr(a, sep=' '): print(sep.join(map(str, a)))
def dd(): return collections.defaultdict(int)
def ddl(): return collections.defaultdict(list)
def ddd(): return collections.defaultdict(collections.defaultdict(int))
 

# region fastio
 
BUFSIZE = 8192
 
 
class FastIO(IOBase):
	newlines = 0
 
	def __init__(self, file):
		self._fd = file.fileno()
		self.buffer = BytesIO()
		self.writable = "x" in file.mode or "r" not in file.mode
		self.write = self.buffer.write if self.writable else None
 
	def read(self):
			while True:
					b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
					if not b:
							break
					ptr = self.buffer.tell()
					self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
			self.newlines = 0
			return self.buffer.read()
 
	def readline(self):
			while self.newlines == 0:
					b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
					self.newlines = b.count(b"\n") + (not b)
					ptr = self.buffer.tell()
					self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
			self.newlines -= 1
			return self.buffer.readline()
 
	def flush(self):
			if self.writable:
					os.write(self._fd, self.buffer.getvalue())
					self.buffer.truncate(0), self.buffer.seek(0)
 
 
class IOWrapper(IOBase):
	def __init__(self, file):
			self.buffer = FastIO(file)
			self.flush = self.buffer.flush
			self.writable = self.buffer.writable
			self.write = lambda s: self.buffer.write(s.encode("ascii"))
			self.read = lambda: self.buffer.read().decode("ascii")
			self.readline = lambda: self.buffer.readline().decode("ascii")
 

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")


for _ in range(ii()):
    n=ii()
    a=li()
    t=0
    for i in range(n-1):
        if a[i]!=a[i+1]:
            t+=1
    if a[-1]!=a[0]:
        t+=1
    if t==0:
        print(1)
        res=[1 for i in range(n)]
        print(*res)
        continue
    if t&1==0:
        print(2)
        res=[0 for i in range(n)]
        f=1
        res[0]=1
        for i in range(1,n):
            if a[i]!=a[i-1]:
                f=3-f
            res[i]=f
        print(*res)
    else:
        if n>t:
            print(2)
            f1=0
            f=1
            res=[0 for i in range(n)]
            res[0]=1
            for i in range(1,n):
                if a[i]==a[i-1]:
                    if f1==0:
                        f1=1
                        f=3-f
                else:
                    f=3-f
                res[i]=f
            print(*res)
        else:
            print(3)
            res=[0 for i in range(n)]
            f=1
            res[0]=1
            for i in range(1,n):
                if a[i]!=a[i-1]:
                    f=3-f
                res[i]=f
            res[-1]=3
            print(*res)