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 

/**
 * GFG
 */
import java.util.*;
import java.io.*;
public class GFG {
static boolean issafe(char ch[][],int i,int j,int dp[][]){
    if(i>=0&&i<ch.length&&j>=0&&j<ch.length&&ch[i][j]!='#'&&dp[i][j]!=1){
        return true;
    }
    return false;
}
static class Edge{
    int first;
    int second;
    Edge(int first,int second){
        this.first=first;
        this.second=second;
    }
}
static void print(ArrayList<Edge>A){
    for(Edge ob:A){
        System.out.println(ob.first+" "+ob.second);
    }
   
}
static void DFS(char ch[][],int i,int j,int dp[][],ArrayList<Edge>A){
int row[]={-1,1,0,0};
int cols[]={0,0,-1,1};
dp[i][j]=1;
A.add(new Edge(i,j));
for(int k=0;k<row.length;++k){
   // System.out.println("Checking");
    if(issafe(ch,i+row[k],j+cols[k],dp)){
       DFS(ch,i+row[k],j+cols[k],dp,A);

    }
           print(A);
  if(A.size()!=0){
      System.out.println();
        A.remove(A.size()-1);
    }
  
  
}

}
static void count(char ch[][]){

    for(int i=0;i<ch.length;++i){
        for(int j=0;j<ch.length;++j){
            int dp[][]=new int[ch.length][ch.length];
            ArrayList<Edge>A=new ArrayList<>();
            if(ch[i][j]=='.'){
                DFS(ch,i,j,dp,A);
            }
        }
    }
}
static final int MAXN = 100001;
      
    // stores smallest prime factor for every number
    static int spf[] = new int[MAXN];
      
    // Calculating SPF (Smallest Prime Factor) for every
    // number till MAXN.
    // Time Complexity : O(nloglogn)
    static void sieve()
    {
        spf[1] = 1;
        for (int i=2; i<MAXN; i++)
      
            // marking smallest prime factor for every
            // number to be itself.
            spf[i] = i;
      
        // separately marking spf for every even
        // number as 2
        for (int i=4; i<MAXN; i+=2)
            spf[i] = 2;
      
        for (int i=3; i*i<MAXN; i++)
        {
            // checking if i is prime
            if (spf[i] == i)
            {
                // marking SPF for all numbers divisible by i
                for (int j=i*i; j<MAXN; j+=i)
      
                    // marking spf[j] if it is not
                    // previously marked
                    if (spf[j]==j)
                        spf[j] = i;
            }
        }
    }
      
    // A O(log n) function returning primefactorization
    // by dividing by smallest prime factor at every step
    static int count=0;
    static Vector<Integer> getFactorization(int x)
    {
        Vector<Integer> ret = new Vector<>();
        while (x != 1)
        {
            ++count;
            ret.add(spf[x]);
            if(spf[x]!=0)
            x = x / spf[x];
        }
        return ret;
 
    }
static int check(String m,String n,int index,String ans,int count){
if(index<0||index>=m.length()){
    return 0;
}
else
if(count<n.length()&&m.charAt(index)!=n.charAt(count)){
    return 0;

}
else 
if(count==n.length()){
    return 1;
}



//we have two choices 4
int first=check(m,n,index-1,ans,count+1);
int second=check(m,n,index+1,ans,count+1);
return first|second;

}
static void print(String txt,int index,String ans,int sum){
if(sum%3==0&&sum!=0){
    System.out.println(ans);
    
}
if(index>=txt.length()){
  
    return;
}
print(txt,index+1,ans+txt.charAt(index),sum+(txt.charAt(index)-48));
print(txt,index+1,ans,sum);

}
public static void main(String[] args) {
Scanner scan=new Scanner(System.in);
int t=scan.nextInt();
while(t-->0){
int n=scan.nextInt();
int m=scan.nextInt();
int arr[]=new int[n];
int dp[]=new int[n+1];
for(int i=0;i<n;++i){
    arr[i]=scan.nextInt();
    dp[arr[i]]++;
}
//Count the total number of ones
int count=0;
int count2=0;
for(int i=0;i<arr.length;++i){
    if(dp[arr[i]]==1){
    ++count;
    }
    else
    if(dp[arr[i]]>=m){++count2;
    dp[arr[i]]=m;

    }
    else {
        dp[arr[i]]=1;
    }
}
int ans=count/m;
int mod=m+1;
int temp=1;
for(int i=0;i<arr.length;++i){
    if(dp[arr[i]]==1&&ans-->0){
       System.out.print(temp+" ");
       ++temp;
       temp=temp%mod;
    }
    else 
    if(dp[arr[i]]!=0){
     System.out.print(Math.min(dp[arr[i]],m)+" ");
     dp[arr[i]]--;
    }
    else{
        System.out.print(0+" ");
    }

}
System.out.println();


}


}}

