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 fastpow(long long int a, long long int p) {
    long long int res = 1;
    while (p) {
      if (p % 2 == 0) {
        a = a * 1ll * a % 1000000007;
        p /= 2;
      } else {
        res = res * 1ll * a % 1000000007;
        p--;
      }
    }
    return res;
  }
  long long int fact(long long int n) {
    long long int res = 1;
    for (long long int i = 1; i <= n; i++) {
      res = res * 1ll * i % 1000000007;
    }
    return res;
  }
  long long int comb(long long int n, long long int k) {
    return fact(n) * 1ll * fastpow(fact(k), 1000000007 - 2) % 1000000007 * 1ll *
           fastpow(fact(n - k), 1000000007 - 2) % 1000000007;
  }
  long long int to_integer(string &aa) {
    long long int res = 0;
    for (auto x : aa) res = (res * 10) + (x - '0');
    return res;
  }
  bool prime(long long int n) {
    if (n <= 1) return false;
    if (n <= 3) return true;
    if (n % 2 == 0 || n % 3 == 0) return false;
    for (long long int i = 5; i * i <= n; i = i + 6)
      if (n % i == 0 || n % (i + 2) == 0) return false;
    return true;
  }
  void sieve(long long int n, vector<long long int> prime) {
    for (long long int p = 2; p * p <= n; p++) {
      if (prime[p] == true) {
        for (long long int i = p * p; i <= n; i += p) prime[i] = false;
      }
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long int tt;
    cin >> tt;
    while (tt--) {
      long long int n, k;
      cin >> n >> k;
      map<long long int, vector<long long int> > pos;
      long long int cnt = 0;
      for (long long int i = 0; i < n; i++) {
        long long int x;
        cin >> x;
        if (pos[x].size() < k) {
          pos[x].push_back(i);
          cnt++;
        }
      }
      cnt -= cnt % k;
      vector<long long int> ans(n);
      for (long long int i = 1; i <= cnt / k; i++) {
        long long int have = 0;
        while (have < k) {
          auto it = pos.begin();
          if (it->second.empty()) {
            pos.erase(it);
            continue;
          }
          have++;
          ans[it->second.back()] = have;
          it->second.pop_back();
        }
      }
      for (auto i : ans) {
        cout << i << ' ';
      }
      cout << "\n";
    }
  }
  