description: 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.
solution: #include <bits/stdc++.h>
  using namespace std;
  struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
      x += 0x9e3779b97f4a7c15;
      x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
      x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
      return x ^ (x >> 31);
    }
    size_t operator()(uint64_t x) const {
      static const uint64_t FIXED_RANDOM =
          chrono::steady_clock::now().time_since_epoch().count();
      return splitmix64(x + FIXED_RANDOM);
    }
  };
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t = 1;
    cin >> t;
    while (t--) {
      int n, k;
      cin >> n >> k;
      int arr[n];
      for (int i = 0; i < n; i++) {
        cin >> arr[i];
      }
      map<int, int> freq;
      for (int i = 0; i < n; i++) {
        freq[arr[i]] += 1;
      }
      vector<pair<int, int>> v;
      map<int, int>::iterator it;
      for (it = freq.begin(); it != freq.end(); it++) {
        v.push_back({it->second, it->first});
      }
      sort(v.begin(), v.end());
      int k_copy = k;
      map<int, queue<int>> ans;
      vector<int> remaining;
      for (int i = 0; i < v.size(); i++) {
        int freqs = v[i].first;
        int num = v[i].second;
        if (freqs >= k) {
          queue<int> s;
          for (int j = 1; j <= k; j++) {
            s.push(j);
          }
          ans[num] = s;
        } else {
          for (int j = 0; j < freqs; j++) {
            remaining.push_back(num);
          }
        }
      }
      vector<int> temp;
      for (int i = 0; i < remaining.size(); i++) {
        if (temp.size() < k) {
          temp.push_back(remaining[i]);
        } else {
          for (int j = 0; j < k; j++) {
            if (ans.find(temp[j]) == ans.end()) {
              queue<int> s;
              s.push(j + 1);
              ans[temp[j]] = s;
            } else {
              ans[temp[j]].push(j + 1);
            }
          }
          temp.clear();
          temp.push_back(remaining[i]);
        }
      }
      if (temp.size() == k) {
        for (int j = 0; j < k; j++) {
          if (ans.find(temp[j]) == ans.end()) {
            queue<int> s;
            s.push(j + 1);
            ans[temp[j]] = s;
          } else {
            ans[temp[j]].push(j + 1);
          }
        }
      }
      for (int i = 0; i < n; i++) {
        if (ans[arr[i]].empty()) {
          cout << 0 << " ";
        } else {
          cout << ans[arr[i]].front() << " ";
          ans[arr[i]].pop();
        }
      }
      cout << (char)(10);
    }
    return 0;
  }
  