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 java.util.*;
import java.io.*;

public class B2 {
    static BufferedReader rd = new BufferedReader(new InputStreamReader(System.in));
    static int mod = 1000000007;
    static StringBuilder sb;
    static int[] ans;

    static void solve() throws Exception {
        String[] input = rd.readLine().split(" ");
        int n = Integer.parseInt(input[0]);
        int colors = Integer.parseInt(input[1]);
        int[] arr = getIntArray();
        ans = new int[n]; // answer array

        // map to store "element" against "indexes of occurrences"
        HashMap<Integer, ArrayList<Integer>> map = new HashMap<>();
        for (int i = 0; i < arr.length; i++) {
            ArrayList<Integer> list;
            if (map.containsKey(arr[i])) {
                list = map.get(arr[i]);
                list.add(i);
                map.put(arr[i], list);
            } else {
                list = new ArrayList<>();
                list.add(i);
                map.put(arr[i], list);
            }
        }

        int color = 1;
        int count = 0;
        for (ArrayList<Integer> list : map.values()) {
            if (list.size() >= colors) { // indexes in list is greater than or equal to "colors"
                for (int i = 0; i < colors; i++) {
                    int index = list.get(i);
                    ans[index] = color;
                    color++;
                    if (color > colors) {
                        color = 1;
                    }
                }
            } else { // indexes less than total number of "colors" available
                count += list.size();
            }
        }

        int rem = count % colors;
        int toBePainted = count - rem; // to paint equal number of elements by colors

        for (ArrayList<Integer> list : map.values()) {
            if (list.size() < colors) {
                for (int i = 0; i < list.size(); i++) {
                    int index = list.get(i);
                    if (toBePainted-- > 0) {
                        ans[index] = color;
                        color++;
                        if (color > colors) {
                            color = 1;
                        }
                    }
                }
            }
        }

        for (int i = 0; i < n; i++) {
            sb.append(ans[i] + " ");
        }
    }

    static class Pair implements Comparable<Pair> {
        int val;
        int idx;

        Pair(int v, int i) {
            val = v;
            idx = i;
        }

        public int compareTo(Pair o) {
            return this.val - o.val;
        }

        @Override
        public int hashCode() {
            int hash = 3;
            hash = 47 * hash + this.val;
            return hash;
        }

        @Override
        public boolean equals(Object obj) {
            if (obj == null) {
                return false;
            }
            if (getClass() != obj.getClass()) {
                return false;
            }
            final Pair other = (Pair) obj;
            if (this.val != other.val) {
                return false;
            }
            return true;
        }
    }

    public static void main(String[] args) throws Exception {
        int test = Integer.parseInt(rd.readLine());
        sb = new StringBuilder();
        while (test-- > 0) {
            solve();
            sb.append("\n");
        }

        System.out.println(sb.toString());
    }

    // FUNCTIONS
    public static void outl(Object str) {
        System.out.println(str);
    }

    public static void out(Object str) {
        System.out.print(str);
    }

    public static int[] getIntArray() throws Exception {
        String[] data = rd.readLine().split(" ");
        int[] arr = new int[data.length];
        for (int i = 0; i < data.length; i++) {
            arr[i] = Integer.parseInt(data[i]);
        }
        return arr;
    }

    public static void displayArray(int[] arr) {
        for (int i = 0; i < arr.length; i++) {
            out(arr[i] + " ");
        }
        outl("");
    }

    public static void displayArray(String[] arr) {
        for (int i = 0; i < arr.length; i++) {
            out(arr[i] + " ");
        }
    }

    // calculates x^n in O(logn), modulo 10^9+7
    static long power(long x, long n) {
        if (n == 0) {
            return 1;
        }

        long xnb2 = power(x, n / 2);
        long xn = (xnb2 * xnb2) % mod;

        if (n % 2 == 1) {
            xn = (xn * x) % mod;
        }
        return xn;
    }
}