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 

# Author Name: Ajay Meena
# Codeforce : https://codeforces.com/profile/majay1638
import sys
import math
import bisect
import heapq
from bisect import bisect_right
from sys import stdin, stdout

# -------------- INPUT FUNCTIONS ------------------


def get_ints_in_variables(): return map(
    int, sys.stdin.readline().strip().split())


def get_int(): return int(sys.stdin.readline())


def get_ints_in_list(): return list(
    map(int, sys.stdin.readline().strip().split()))
def get_list_of_list(n): return [list(
    map(int, sys.stdin.readline().strip().split())) for _ in range(n)]


def get_string(): return sys.stdin.readline().strip()

# -------- SOME CUSTOMIZED FUNCTIONS-----------


def myceil(x, y): return (x + y - 1) // y

# -------------- SOLUTION FUNCTION ------------------


def Solution(arr, n, k):
    # Write Your Code Here
    hm = {}
    for i in range(n):
        if arr[i] in hm:
            hm[arr[i]].append(i)
        else:
            hm[arr[i]] = [i]
    res = [0 for _ in range(n)]
    tmp = []
    for el in hm:
        if len(hm[el]) >= k:
            mn = k
            while mn:
                res[hm[el].pop()] = mn
                mn -= 1
        else:
            for v in hm[el]:
                tmp.append(v)
    for i in range(0, (len(tmp)-k)+1, k):
        for j in range(k):
            res[tmp[i+j]] = j+1
    print(*res)


def main():
    # Take input Here and Call solution function
    for _ in range(get_int()):
        n, k = get_ints_in_variables()
        arr = get_ints_in_list()
        Solution(arr, n, k)


# calling main Function
if __name__ == '__main__':
    main()
