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 

// package com.company

/*
 * @author :: Yuvraj Singh
 * CS UNDERGRAD AT IT GGV BILASPUR
 */

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;

import static java.lang.Math.*;
import static java.lang.System.out;
import static java.lang.Long.MAX_VALUE;

public final class Main {

    FastReader in;
    StringBuffer sb;
    public static void main(String[] args) {
        new Main().run();
    }

    void run(){
        in= new FastReader();
        start();
    }

    void start(){
        sb= new StringBuffer();
        for(int t=in.nextInt();t>0;t--)
            solve();
        out.print(sb);
    }

    void solve(){
        int n= in.nextInt();
        int k= in.nextInt();
        int[] arr = new int[n];
        for(int i=0;i<n;i++)
            arr[i]= in.nextInt();
        HashMap<Integer,Integer> mp = new HashMap<>();
        for(int i : arr) {
            mp.put(i,mp.getOrDefault(i,0)+1);
        }
        Set<Integer> x = mp.keySet();
        HashMap<Integer,Integer> xk = new HashMap<>();
        HashMap<Integer,Integer> pk = new HashMap<>();
        int cnt = 0;
        for(int i : x) {
            int y = mp.get(i);
            if(y>=k) {
                xk.put(i,k);
            }
            else {
                pk.put(i,y);
                cnt+=y;
            }
        }
        int h = cnt%k;
        cnt-=h;
        HashMap<Integer,Stack<Integer> > hk = new HashMap<>();
        for(int i : pk.keySet()) {
            hk.put(i,new Stack<Integer>());
            hk.get(i).add(0);
        }
        for(int i : pk.keySet()) {
            if(cnt>0) {
                int y = pk.get(i);
                while( cnt > 0&&y>0) {
                    hk.get(i).add((cnt%k)+1);
                    y--;
                    cnt--;
                }
            }
            else
                break;
        }
        StringBuffer ans = new StringBuffer();
        for(int i : arr) {
            if(xk.containsKey(i)) {
                int j = xk.get(i);
                if(j>0) {
                    ans.append(j+" ");
                    xk.put(i,j-1);
                }
                else
                    ans.append(0+" ");
            }
            else {
                if(hk.get(i).peek() != 0) {
                    int ha =hk.get(i).pop();
                    ans.append(ha+" ");
                }
                else
                    ans.append(0+" ");
            }
        }
        sb.append(ans).append("\n");
    }

    long power(long x, long y, long p) {
        long res = 1;
        x = x % p;
        while (y > 0) {
            if (y % 2 == 1)
                res = (res * x) % p;

            y = y >> 1;
            x = (x * x) % p;
        }
        return res;
    }

    int upper_bound(long[] arr, int key) {
        int i=0, j=arr.length-1;
        if (arr[j]<=key) return j+1;
        if(arr[i]>key) return i;
        while (i<j){
            int mid= (i+j)/2;
            if(arr[mid]<=key){
                i= mid+1;
            }else{
                j=mid;
            }
        }
        return i;
    }

    void sort(long[] A){
        int n = A.length;
        Random rnd = new Random();
        for(int i=0; i<n; ++i){
            long tmp = A[i];
            int randomPos = i + rnd.nextInt(n-i);
            A[i] = A[randomPos];
            A[randomPos] = tmp;
        }
        Arrays.sort(A);
    }

    long[] longArr(int n){
        long[] res= new long[n];
        for(int i=0;i<n;i++){
            res[i]= in.nextLong();
        }
        return res;
    }

    // sieve of eratosthenes code for precomputing whether numbers are prime or not up to MAX_VALUE
    long MAX= MAX_VALUE;
    int[] precomp= new int[(int) (MAX+1)];
    void sieve(){
        long n= MAX;
        boolean[] prime = new boolean[(int) (n+1)];
        for(int i=0;i<=n;i++)
            prime[i] = true;

        for(long p = 2; p*p <=n; p++)
        {
            // If prime[p] is not changed, then it is a prime
            if(prime[(int) p])
            {
                // Update all multiples of p
                for(long i = p*p; i <= n; i += p)
                    prime[(int) i] = false;
            }
        }

        // Print all prime numbers
        for(long i = 2; i <= n; i++)
        {
            if(prime[(int) i])
                precomp[(int) i]= 1;
        }
    }

    boolean isDigitSumPalindrome(long N) {
        // code here
        long sum= sumOfDigits(String.valueOf(N));
        long rev=0;
        long org= sum;
        while (sum!=0){
            long d= sum%10;
            rev = rev*10 +d;
            sum /= 10;
        }
        return org == rev;
    }

    long sumOfDigits(String n){
        long sum= 0;
        for (char c: n.toCharArray()){
            sum += Integer.parseInt(String.valueOf(c));
        }
        return sum;
    }

    long[] revArray(long[] arr) {
        int n= arr.length;
        int i=0, j=n-1;
        while (i<j){
            long temp= arr[i];
            arr[i]= arr[j];
            arr[j]= temp;
            i++;
            j--;
        }
        return arr;
    }

    long gcd(long a, long b){
        if (b==0)
            return a;
        return gcd(b, a%b);
    }
    long lcm(long a,long b){
        return (a*b)/gcd(a,b);
    }

    static class Pair{
        long first;
        long second;
        Pair(long x,long y){
            this.first=x;
            this.second=y;
        }
    }

    static class Compare {
        static void compare(ArrayList<Pair> arr, long n) {
            arr.sort(new Comparator<Pair>() {
                @Override
                public int compare(Pair p1, Pair p2) {
                    return (int) (p2.first - p1.first);
                }
            });
        }
    }

    public static class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br=new BufferedReader(new InputStreamReader(System.in));
        }
        String next(){
            while (st==null || !st.hasMoreElements()){
                try{
                    st=new StringTokenizer(br.readLine());
                }catch (Exception e){
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt(){
            return Integer.parseInt(next());
        }
        long nextLong(){
            return Long.parseLong(next());
        }
        double nextDouble(){
            return Double.parseDouble(next());
        }
        float nextFloat(){
            return Float.parseFloat(next());
        }
        String nextLine(){
            String str="";
            try{
                str=br.readLine();
            }catch (Exception e){
                e.printStackTrace();
            }
            return str;
        }
    }
}