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 <typename T>
  T gcd(T a, T b) {
    return b ? gcd(b, a % b) : a;
  }
  template <typename T>
  T lcm(T a, T b) {
    return a * b / gcd(a, b);
  }
  const int mod = 998244353;
  const int MOD = 1e9 + 7;
  inline long long read() {
    long long ans = 0;
    char ch = getchar(), las = ' ';
    while (!isdigit(ch)) las = ch, ch = getchar();
    while (isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
    if (las == '-') ans = -ans;
    return ans;
  }
  inline void write(long long x) {
    if (x < 0) x = -x, putchar('-');
    if (x >= 10) write(x / 10);
    putchar(x % 10 + '0');
  }
  const int N = 4e5;
  int b[N];
  void run_case() {
    int n = read(), k = read();
    map<int, int> m;
    map<int, deque<int> > g;
    for (int i = (1); i <= (n); i++) {
      b[i] = read();
      m[b[i]]++;
    }
    int f = 0;
    map<int, int>::iterator it;
    for (it = m.begin(); it != m.end(); it++) {
      if (it->second == k) {
        for (int j = 1; j <= it->second; j++) {
          g[it->first].push_back((j % k == 0) ? k : j % k);
        }
      } else if (it->second > k) {
        for (int j = 1; j <= it->second; j++) {
          if (j <= k)
            g[it->first].push_back((j % k == 0) ? k : j % k);
          else
            g[it->first].push_back(0);
        }
      } else {
        for (int j = 1; j <= it->second; j++) {
          f++;
          g[it->first].push_back((f % k == 0) ? k : f % k);
        }
      }
    }
    int ff, tt;
    map<int, int>::reverse_iterator iter;
    if (f % k) {
      for (iter = m.rbegin(); iter != m.rend(); iter++) {
        if (iter->second < k) {
          ff = f;
          for (int j = 1; j <= iter->second; j++) {
            f--;
            g[iter->first].pop_back();
            if (f % k == 0) {
              tt = iter->first;
              goto end;
            }
          }
          for (int j = 1; j <= iter->second; j++) {
            g[iter->first].push_back(0);
          }
        }
      }
    end:
      for (int i = 1; i <= ff - f; i++) g[tt].push_back(0);
    }
    for (int i = 1; i <= n; i++) {
      if (i != 1) printf(" ");
      printf("%d", g[b[i]].front());
      g[b[i]].pop_front();
    }
    puts("");
  }
  int main() {
    int _ = read();
    while (_--) {
      run_case();
    }
  }
  