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 t;
  long long dd[200005];
  struct hdz {
    long long i, j;
  };
  hdz A[200005];
  bool cmp(hdz a, hdz b) { return a.j < b.j; }
  int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cin >> t;
    while (t--) {
      long long n, k;
      cin >> n >> k;
      long long a[n + 2];
      map<long long, long long> m;
      vector<long long> v[n + 2];
      set<long long> s;
      for (long long i = 1; i <= n; i++) {
        dd[i] = 0;
        A[i].i = 0;
        A[i].j = 0;
        cin >> a[i];
        m[a[i]]++;
        v[a[i]].push_back(i);
        s.insert(a[i]);
      }
      long long tmp = 0;
      for (long long i : s) {
        A[++tmp].i = i;
        A[tmp].j = m[i];
      }
      if (k == n) {
        for (long long i = 1; i <= n; i++) {
          cout << i << ' ';
        }
        cout << '\n';
        continue;
      }
      sort(A + 1, A + 1 + tmp, cmp);
      for (long long i = tmp; i >= 1; i--) {
        long long tmp = k;
        if (tmp > A[i].j) {
          long long d = 0;
          long long res = 0;
          bool kt = false;
          for (long long j = i; j >= 1; j--) {
            res += A[j].j;
            if (res >= tmp) {
              kt = true;
              break;
            }
          }
          long long j = i;
          if (kt)
            while (1) {
              if (tmp > A[j].j) {
                tmp -= A[j].j;
                A[j].j = 0;
                for (long long k = 0; k < v[A[j].i].size(); k++) {
                  long long f = v[A[j].i][k];
                  if (dd[f] != 0) continue;
                  dd[f] = 1;
                  a[f] = ++d;
                }
              } else {
                A[j].j = A[j].j - tmp;
                for (long long q = 0; q < v[A[j].i].size(); q++) {
                  long long f = v[A[j].i][q];
                  if (dd[f] != 0) continue;
                  if (d == k) break;
                  dd[f] = 1;
                  a[f] = ++d;
                }
                cout << "\n\n";
                break;
              }
              j--;
            }
        } else {
          long long d = 0;
          for (long long j = 0; j < (long long)v[A[i].i].size(); j++) {
            if (d == k) break;
            long long f = v[A[i].i][j];
            if (dd[f] != 0) continue;
            dd[f] = 1;
            a[f] = ++d;
          }
        }
      }
      for (long long i = 1; i <= n; i++) {
        if (dd[i] != 0)
          cout << a[i] << ' ';
        else
          cout << 0 << ' ';
      }
      cout << '\n';
    }
  }
  