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 

#include <bits/stdc++.h>
using namespace std;
bool secCompare(pair<int, int> a, pair<int, int> b) {
  if (a.second < b.second) {
    return true;
  }
  return false;
}
int main() {
  int T;
  cin >> T;
  while (T > 0) {
    int N, K;
    cin >> N >> K;
    vector<int> V;
    vector<pair<int, int>> v;
    unordered_map<int, int> umap;
    int x;
    for (int i = 0; i < N; i++) {
      cin >> x;
      v.push_back({x, i});
      if (umap.find(x) == umap.end()) {
        umap[x] = 1;
        V.push_back(x);
      } else {
        umap[x]++;
      }
    }
    int total = 0;
    for (int i = 0; i < V.size(); i++) {
      total += min(umap[V[i]], K);
    }
    int maxC = total / K;
    sort(v.begin(), v.end());
    int prev = -1;
    int cur = 1;
    int last = 1;
    int count = 0;
    for (int i = 0; i < v.size(); i++) {
      if (prev == -1) {
        prev = v[i].first;
        cur = 1;
        if (last == 1) {
          count++;
        }
        if (count > maxC) {
          last = 0;
        }
        v[i].first = last;
        if (last != 0) {
          last++;
        }
        if (last > K) {
          last = 1;
        }
      } else if (v[i].first == prev) {
        cur++;
        if (cur <= K) {
          if (last == 1) {
            count++;
          }
          if (count > maxC) {
            last = 0;
          }
          v[i].first = last;
          if (last != 0) {
            last++;
          }
          if (last > K) {
            last = 1;
          }
        } else {
          v[i].first = 0;
        }
      } else {
        prev = v[i].first;
        cur = 1;
        if (last == 1) {
          count++;
        }
        if (count > maxC) {
          last = 0;
        }
        v[i].first = last;
        if (last != 0) {
          last++;
        }
        if (last > K) {
          last = 1;
        }
      }
    }
    sort(v.begin(), v.end(), secCompare);
    for (int i = 0; i < v.size(); i++) {
      cout << v[i].first << " ";
    }
    cout << "\n";
    T--;
  }
}
