This problem is an extension of the problem "Wonderful Coloring - 1". It has quite many differences, so you should read this statement completely.

Recently, Paul and Mary have found a new favorite sequence of integers a_1, a_2, ..., a_n. They want to paint it using pieces of chalk of k colors. The coloring of a sequence is called wonderful if the following conditions are met:

  1. each element of the sequence is either painted in one of k colors or isn't painted; 
  2. each two elements which are painted in the same color are different (i. e. there's no two equal values painted in the same color); 
  3. let's calculate for each of k colors the number of elements painted in the color — all calculated numbers must be equal; 
  4. the total number of painted elements of the sequence is the maximum among all colorings of the sequence which meet the first three conditions. 



E. g. consider a sequence a=[3, 1, 1, 1, 1, 10, 3, 10, 10, 2] and k=3. One of the wonderful colorings of the sequence is shown in the figure.

<image> The example of a wonderful coloring of the sequence a=[3, 1, 1, 1, 1, 10, 3, 10, 10, 2] and k=3. Note that one of the elements isn't painted.

Help Paul and Mary to find a wonderful coloring of a given sequence a.

Input

The first line contains one integer t (1 ≤ t ≤ 10000) — the number of test cases. Then t test cases follow.

Each test case consists of two lines. The first one contains two integers n and k (1 ≤ n ≤ 2⋅10^5, 1 ≤ k ≤ n) — the length of a given sequence and the number of colors, respectively. The second one contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n).

It is guaranteed that the sum of n over all test cases doesn't exceed 2 ⋅ 10^5.

Output

Output t lines, each of them must contain a description of a wonderful coloring for the corresponding test case.

Each wonderful coloring must be printed as a sequence of n integers c_1, c_2, ..., c_n (0 ≤ c_i ≤ k) separated by spaces where

  * c_i=0, if i-th element isn't painted; 
  * c_i>0, if i-th element is painted in the c_i-th color. 



Remember that you need to maximize the total count of painted elements for the wonderful coloring. If there are multiple solutions, print any one.

Example

Input


6
10 3
3 1 1 1 1 10 3 10 10 2
4 4
1 1 1 1
1 1
1
13 1
3 1 4 1 5 9 2 6 5 3 5 8 9
13 2
3 1 4 1 5 9 2 6 5 3 5 8 9
13 3
3 1 4 1 5 9 2 6 5 3 5 8 9


Output


1 1 0 2 3 2 2 1 3 3
4 2 1 3
1
0 0 1 1 0 1 1 1 0 1 1 1 0
2 1 2 2 1 1 1 1 2 1 0 2 2
1 1 3 2 1 3 3 1 2 2 3 2 0

Note

In the first test case, the answer is shown in the figure in the statement. The red color has number 1, the blue color — 2, the green — 3.
12
10 3
3 1 1 1 1 10 3 10 10 2
4 4
1 1 1 1
1 1
1
13 1
3 1 4 1 5 9 2 6 5 3 5 8 9
13 2
3 1 4 1 5 9 2 6 5 3 5 8 9
13 3
3 1 4 1 5 9 2 6 5 3 5 8 9
10 3
3 1 1 1 1 10 3 10 10 2
4 4
1 1 1 1
1 1
1
13 1
3 1 4 1 5 9 2 6 5 3 5 8 9
13 2
3 1 4 1 5 9 2 6 5 3 5 8 9
13 3
3 1 4 1 5 9 2 6 5 3 5 8 9
 6
10 3
3 1 1 1 1 10 3 10 10 2
4 4
1 1 1 1
1 1
1
13 1
3 1 4 1 5 9 2 6 5 3 5 8 9
13 2
3 1 4 1 5 9 2 6 5 3 5 8 9
13 3
3 1 4 1 5 9 2 6 5 3 5 8 9

2 1 2 3 0 1 3 2 3 1 
1 2 3 4 
1 
1 1 1 0 1 1 1 1 0 0 0 1 0 
2 1 2 2 1 1 1 1 2 1 0 2 2 
1 1 3 2 1 3 3 1 2 2 3 2 0 
2 1 2 3 0 1 3 2 3 1 
1 2 3 4 
1 
1 1 1 0 1 1 1 1 0 0 0 1 0 
2 1 2 2 1 1 1 1 2 1 0 2 2 
1 1 3 2 1 3 3 1 2 2 3 2 0 
 2 1 2 3 0 1 3 2 3 1 
1 2 3 4 
1 
1 1 1 0 1 1 1 1 0 0 0 1 0 
2 1 2 2 1 1 1 1 2 1 0 2 2 
1 1 3 2 1 3 3 1 2 2 3 2 0 

import os
import sys
from io import BytesIO, IOBase
import math
from queue import Queue
import collections
import itertools
import bisect
import heapq
# sys.setrecursionlimit(100000)
# ^^^TAKE CARE FOR MEMORY LIMIT^^^
import random


def main():
    pass


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")


def binary(n):
    return (bin(n).replace("0b", ""))


def decimal(s):
    return (int(s, 2))


def pow2(n):
    p = 0
    while (n > 1):
        n //= 2
        p += 1
    return (p)


def primeFactors(n):
    cnt = []
    while n % 2 == 0:
        cnt.append(2)
        n = n / 2
    for i in range(3, int(math.sqrt(n)) + 1, 2):
        while n % i == 0:
            cnt.append(i)
            n = n / i
    if n > 2:
        cnt.append(int(n))
    return (cnt)

def primeFactorsCount(n):
    cnt=0
    while n % 2 == 0:
        cnt+=1
        n = n // 2
    for i in range(3, int(math.sqrt(n)) + 1, 2):
        while n % i == 0:
            cnt+=1
            n = n // i
    if n > 2:
        cnt+=1
    return (cnt)

def isPrime(n):
    if (n == 1):
        return (False)
    else:
        root = int(n ** 0.5)
        root += 1
        for i in range(2, root):
            if (n % i == 0):
                return (False)
        return (True)


def maxPrimeFactors(n):
    maxPrime = -1
    while n % 2 == 0:
        maxPrime = 2
        n >>= 1
    for i in range(3, int(math.sqrt(n)) + 1, 2):
        while n % i == 0:
            maxPrime = i
            n = n / i
    if n > 2:
        maxPrime = n
    return int(maxPrime)


def countcon(s, i):
    c = 0
    ch = s[i]
    for i in range(i, len(s)):
        if (s[i] == ch):
            c += 1
        else:
            break
    return (c)


def lis(arr):
    n = len(arr)
    lis = [1] * n
    for i in range(1, n):
        for j in range(0, i):
            if arr[i] > arr[j] and lis[i] < lis[j] + 1:
                lis[i] = lis[j] + 1
    maximum = 0
    for i in range(n):
        maximum = max(maximum, lis[i])
    return maximum


def isSubSequence(str1, str2):
    m = len(str1)
    n = len(str2)
    j = 0
    i = 0
    while j < m and i < n:
        if str1[j] == str2[i]:
            j = j + 1
        i = i + 1
    return j == m


def maxfac(n):
    root = int(n ** 0.5)
    for i in range(2, root + 1):
        if (n % i == 0):
            return (n // i)
    return (n)


def p2(n):
    c = 0
    while (n % 2 == 0):
        n //= 2
        c += 1
    return c


def seive(n):
    primes = [True] * (n + 1)
    primes[1] = primes[0] = False
    i = 2
    while (i * i <= n):
        if (primes[i] == True):
            for j in range(i * i, n + 1, i):
                primes[j] = False
        i += 1
    pr = []
    for i in range(0, n + 1):
        if (primes[i]):
            pr.append(i)
    return pr


def ncr(n, r, p):
    num = den = 1
    for i in range(r):
        num = (num * (n - i)) % p
        den = (den * (i + 1)) % p
    return (num * pow(den,
                      p - 2, p)) % p


def denofactinverse(n, m):
    fac = 1
    for i in range(1, n + 1):
        fac = (fac * i) % m
    return (pow(fac, m - 2, m))


def numofact(n, m):
    fac = 1
    for i in range(1, n + 1):
        fac = (fac * i) % m
    return (fac)

def sod(n):
    s = 0
    while (n > 0):
        s += n % 10
        n //= 10
    return s
'''for xyz in range(0,int(input())):
    n,k=map(int,input().split())
    l=list(map(int,input().split()))
    cnt=[0]*(max(l)+1)
    for i in l:
        cnt[i]+=1
    #print(cnt)
    val=0
    for i in cnt:
        val+=min(i,k)
    #print(val)
    maxC=val//k
    #print(maxC)
    exc=n-maxC*k
    tbv=[0]*(1+max(l))
    for i in range(0,len(cnt)):
        tbv[i]=min(cnt[i],k)
        exc-=cnt[i]-tbv[i]
    print(exc)
    prevcnt=0
    prevCol=0
    col=[1]*(len(cnt))
    for i in set(l):
        currCol=((prevcnt+prevCol)%k)+1
        col[i]=currCol
        prevCol=currCol
        prevcnt=tbv[i]
    #print(col)
    ans=[]
    for i in range(0,n):
        val=l[i]
        if(tbv[val]==0):
            ans.append(0)
        else:
            ans.append(col[val])
            col[val]=(col[val])%(k)+1
            tbv[val]-=1
    print(*ans)'''
for xyz in range(0,int(input())):
    n,k=map(int,input().split())
    l=list(map(int,input().split()))
    MA=max(l)
    pos=[[] for i in range(0,MA+1)]
    tc=0
    for i in range(0,n):
        if(len(pos[l[i]])<k):
            pos[l[i]].append(i)
            tc+=1
    col=[[]for i in range(0,k)]
    trim=tc%k
    cc=0
    for i in range(0,len(pos)):
        cl=pos[i]
        for j in range(0,len(cl)):
            if(trim>0):
                trim-=1
                continue
            col[cc].append(pos[i][j])
            cc=(cc+1)%k
    ans=[0]*n
    for i in range(0,len(col)):
        cuco=i+1
        cl=col[i]
        for j in cl:
            ans[j]=cuco
    print(*ans)