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

public class CodeForces {
    /*****************************************WRITING CODE STARTS HERE******************************************/

    public static void solve(int tCase)throws IOException {
            int n = sc.nextInt();
            int k = sc.nextInt();
            ArrayList<Integer> [] arr = new ArrayList[n+1];
            for(int i = 1;i<=n;i++)arr[i] = new ArrayList<>();
            int[] ans = new int[n];
            for(int i=0;i<n;i++){
                int x = sc.nextInt();
                arr[x].add(i);
            }
            int totalColored = 0;
            for(int i=1;i<=n;i++){
                totalColored += Math.min(arr[i].size(),k);
            }
            int colored = 0;
            int color = 1;
            for(int i=1;i<=n;i++){
                for(int j =0;j<Math.min(arr[i].size(),k);j++){
                    ans[arr[i].get(j)] = color++;
                    if(color == k+1)color = 1;
                    colored++;
                    if(colored == totalColored)break;
                }
            }
            for(int ele : ans)out.print(ele+" ");
            out.println();


    }

    /*****************************************WRITING CODE ENDS HERE******************************************/

    public static void main(String[] args) throws IOException{
        int testCase = 1;
        testCase = sc.nextInt();
        for (int i = 1; i <= testCase; i++) solve(i);
        closeIO();
    }
    static FastestReader sc = new FastestReader();              // FASTEST INPUT
    static PrintWriter out = new PrintWriter(System.out);                // FAST OUTPUT


    /**************************************HELPER FUNCTION STARTS HERE ************************************/

    public static int mod = (int) 1e9 + 7;
    public static final int inf_int = (int) 1e9+10;
    public static final long inf_long = (long) 1e15;

    public static final String YES = "YES";
    public static final String NO = "NO";

    // euclidean algorithm time O(max (loga ,logb))
    public static long gcd(long a, long b) {
        long x = Math.min(a, b);
        long y = Math.max(a, b);
        if (y % x == 0) return x;
        return gcd(y % x, x);
    }

    public static long lcm(long a, long b) {
        // lcm(a,b) * gcd(a,b) = a * b
        return (a / gcd(a, b)) * b;
    }
    // binary exponentiation time O(logn)
    public static long power(long x, long n) {
        long ans = 1;
        while (n > 0) {
            if ((n & 1) == 1) {
                ans *= x;
                ans %= mod;
                n--;
            } else {
                x *= x;
                x %= mod;
                n >>= 1;
            }
        }
        return ans;
    }

    public static long firstDivisor(long n) {
        if (n == 1 || n == 0) return n;
        for (long i = 2; i * i <= n; i++)
            if (n % i == 0) return i;
        return -1;
    }


    /**************************************HELPER FUNCTION ENDS HERE ************************************/

    /*****************************************FAST INPUT STARTS HERE *************************************/


    public static void closeIO() throws IOException{
        out.flush();
        out.close();
        sc.close();
    }
    private static class FastestReader {
        private static final int BUFFER_SIZE = 1 << 16;
        private final DataInputStream din;
        private final byte[] buffer;
        private int bufferPointer, bytesRead;

        public FastestReader() {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public FastestReader(String file_name) throws IOException {
            din = new DataInputStream(new FileInputStream(file_name));
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        private static boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }

        private int skip() throws IOException {
            int b;
            //noinspection StatementWithEmptyBody
            while ((b = read()) != -1 && isSpaceChar(b)) {}
            return b;
        }

        public String next() throws IOException {
            int b = skip();
            final StringBuilder sb = new StringBuilder();
            while (!isSpaceChar(b)) { // when nextLine, (isSpaceChar(b) && b != ' ')
                sb.appendCodePoint(b);
                b = read();
            }
            return sb.toString();
        }

        public int nextInt() throws IOException {
            int ret = 0;
            byte c = read();
            while (c <= ' ') {
                c = read();
            }
            final boolean neg = c == '-';
            if (neg) { c = read(); }
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');

            if (neg) { return -ret; }
            return ret;
        }

        public long nextLong() throws IOException {
            long ret = 0;
            byte c = read();
            while (c <= ' ') { c = read(); }
            final boolean neg = c == '-';
            if (neg) { c = read(); }
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');
            if (neg) { return -ret; }
            return ret;
        }

        public double nextDouble() throws IOException {
            double ret = 0, div = 1;
            byte c = read();
            while (c <= ' ') { c = read(); }
            final boolean neg = c == '-';
            if (neg) { c = read(); }

            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');

            if (c == '.') {
                while ((c = read()) >= '0' && c <= '9') {
                    ret += (c - '0') / (div *= 10);
                }
            }

            if (neg) { return -ret; }
            return ret;
        }

        private void fillBuffer() throws IOException {
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
            if (bytesRead == -1) { buffer[0] = -1; }
        }

        private byte read() throws IOException {
            if (bufferPointer == bytesRead) { fillBuffer(); }
            return buffer[bufferPointer++];
        }

        public void close() throws IOException {
            din.close();
        }

    }
    /*****************************************FAST INPUT ENDS HERE *************************************/
}