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;
  template <class T>
  void fr(T &a, bool f = 0, char ch = getchar()) {
    for (a = 0; ch < '0' || ch > '9'; ch = getchar()) ch == '-' ? f = 1 : 0;
    for (; ch >= '0' && ch <= '9'; ch = getchar()) a = a * 10 + ch - '0';
    a = f ? -a : a;
  }
  template <class T, class... Y>
  void fr(T &t, Y &...a) {
    fr(t), fr(a...);
  }
  int fr() {
    int a;
    return fr(a), a;
  }
  const int N = 2e5;
  int a[N + 10], cnt[N + 10], ans[N + 10];
  vector<int> p[N + 10];
  struct OI {
    int RP, score;
  } CSPS2021, NOIP2021;
  int main() {
    CSPS2021.RP++, CSPS2021.score++, NOIP2021.RP++, NOIP2021.score++, 392699;
    for (int T = fr(); T--;) {
      int n = fr(), k = fr();
      ans[0] = 0;
      int pos = 0;
      for (int i = 1; i <= n; i++) fr(a[i]), cnt[a[i]]++, p[a[i]].push_back(i);
      for (int i = 1; i <= n; i++)
        if (cnt[i] >= k) {
          for (int j = 0; j < k; j++) ans[p[i][j]] = j + 1;
        } else
          ans[0] += cnt[i];
      ans[0] = (ans[0] / k) * k;
      for (int i = 1; i <= n; i++) {
        if (cnt[i] >= k) continue;
        if (ans[0] == 0) continue;
        for (int j = 0; j < cnt[i]; j++)
          if (ans[0]) ans[p[i][j]] = pos % k + 1, pos = ans[p[i][j]], ans[0]--;
      }
      for (int i = 1; i <= n; i++)
        printf("%d ", ans[i]), cnt[i] = 0, ans[i] = 0, p[i].clear();
      puts("");
    }
    return 0;
  }
  