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 sys
testing = len(sys.argv) == 4 and sys.argv[3] == "myTest"
interactive = False
if testing:
    cmd = sys.stdout
    from time import time
    start_time = int(round(time() * 1000)) 
    readAll = open(sys.argv[1], 'r').read
    sys.stdout = open(sys.argv[2], 'w')
else:
    readAll = sys.stdin.read

# ############ ---- I/O Functions ---- ############

class InputData:
    def __init__(self):
        self.lines = readAll().split('\n')
        self.n = len(self.lines)
        self.ii = -1
    def input(self):
        self.ii += 1
        assert self.ii < self.n
        return self.lines[self.ii]

flush = sys.stdout.flush
if interactive and not testing:
    input = sys.stdin.readline
else:
    inputData = InputData()
    input = inputData.input

def intin():
    return(int(input()))
def intlin():
    return(list(map(int,input().split())))
def chrin():
    return(list(input()))
def strin():
    return input()
def lout(l, sep="\n", toStr=True):
    print(sep.join(map(str, l) if toStr else l))
def dout(*args, **kargs):
    if not testing: return
    if args: print(args[0] if len(args)==1 else args)
    if kargs: print([(k,v) for k,v in kargs.items()])
def ask(q):
    sys.stdout.write(str(q)+'\n')
    flush()
    return intin()
    
# ############ ---- I/O Functions ---- ############

# from math import ceil
from collections import defaultdict as ddict, Counter
# from heapq import *
# from Queue import Queue

def main():
    n,k = intlin()
    a = intlin()
    cmap = ddict(list)
    for i in xrange(n):
        cmap[a[i]].append(i)
    cnts = Counter(a)
    cnt = 0
    ans = [0]*n
    j = 0
    tmp = []
    for x in cmap:
        if len(cmap[x]) >= k:
            for i in xrange(k):
                ans[cmap[x][i]] = i
        else:
            for i in xrange(len(cmap[x])):
                ans[cmap[x][i]] = j
                tmp.append(cmap[x][i])
                if len(tmp) == k:
                    tmp = []
                j += 1
                j %= k
    ans = [x+1 for x in ans]
    for i in tmp:
        ans[i] = 0
    return ' '.join(map(str,ans))


anss = []
for _ in xrange(intin()):
    anss.append(main())
    # anss.append("YES" if main() else "NO")
lout(anss)

if testing:
    sys.stdout = cmd
    print(int(round(time() * 1000))  - start_time)