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;
  long long int fp(long long int n, long long int p) {
    if (p == 0) return 1;
    if (p == 1) return n;
    long long int res = fp(n, p / 2) % (long long)(1e9 + 7);
    res = ((res) * (res)) % (long long)(1e9 + 7);
    if (p % 2) res = (res * n) % (long long)(1e9 + 7);
    return res;
  }
  string tostring(long long int a) {
    string s = "";
    while (a > 0) {
      char c = a % 10 + '0';
      s.push_back(c);
      a /= 10;
    }
    reverse(s.begin(), s.end());
    return s;
  }
  int t, n, k, x, arr[200005];
  int main() {
    ios::sync_with_stdio(0);
    ios_base::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
    ;
    cin >> t;
    while (t--) {
      cin >> n >> k;
      vector<vector<int>> v(n + 5);
      memset(arr, -1, sizeof(arr));
      map<int, int> mp;
      int cnt = 0, ind = 0, rem = 0, full = 0, c = 0;
      for (int i = 0; i < n; i++) {
        cin >> x;
        mp[x]++;
        v[x].push_back(i);
        if (mp[x] == k)
          full++;
        else if (mp[x] > k)
          rem++, arr[i] = -1;
      }
      c = n - (full * k) - rem;
      c = c - (c % k);
      for (auto i : mp) {
        if (i.second >= k) {
          for (int j = 0; j < k; j++) {
            arr[v[i.first][j]] = j + 1;
          }
        } else {
          for (int j = 0; j < v[i.first].size() && c > 0; j++) {
            arr[v[i.first][j]] = cnt + 1;
            cnt = (cnt + 1) % k;
            c--;
          }
        }
      }
      for (int i = 0; i < n; i++)
        if (arr[i] == -1)
          cout << 0 << " ";
        else
          cout << arr[i] << " ";
      cout << '\n';
    }
    return 0;
  }
  