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;
  const long long mod = 1e9 + 17;
  const long long inf = 1e18 + 1;
  long long modi(long long numberToMod) {
    numberToMod += mod;
    numberToMod %= mod;
    return numberToMod;
  }
  long long powerWithMod(long long numberToPower, long long myPower) {
    if (!myPower) return long long(1);
    long long recursive = powerWithMod(numberToPower, myPower / long long(2));
    if (myPower % 2) return modi(modi(recursive * recursive) * numberToPower);
    return modi(recursive * recursive);
  }
  long long power(long long numberToPower, long long myPower) {
    if (!myPower) return long long(1);
    long long recursive = powerWithMod(numberToPower, myPower / long long(2));
    if (myPower % 2) return recursive * recursive * numberToPower;
    return recursive * recursive;
  }
  long long gcd(long long numberOne, long long numberTwo) {
    if (numberTwo == 0) return numberOne;
    return gcd(numberTwo, numberOne % numberTwo);
  }
  vector<long long> compute_pi(string pattern) {
    vector<long long> result(pattern.size());
    result[0] = 0;
    for (unsigned int i = 1; i < pattern.size(); i++) {
      result[i] = 0;
      int KMP_count = result[i - 1];
      while (KMP_count > 0 && pattern[KMP_count] != pattern[i])
        KMP_count = result[KMP_count - 1];
      if (pattern[KMP_count] == pattern[i]) result[i] = KMP_count + 1;
    }
    return result;
  }
  const int maxn = 2e5 + 10;
  long long t;
  long long m, x, k, a[maxn], n, c[maxn];
  vector<int> v[maxn], save;
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> t;
    while (t--) {
      cin >> n >> k;
      for (int i = 1; i <= n; i++) v[i].clear();
      for (int i = 0; i < n; i++) c[i] = 0;
      for (int i = 0; i < n; i++) {
        cin >> a[i];
        v[a[i]].push_back(i);
      }
      m = 0;
      x = 0;
      save.clear();
      for (int i = 1; i <= n; i++) {
        if (v[i].size() >= k) {
          for (int i2 = 0; i2 < k; i2++) c[v[i][i2]] = i2 + 1;
        } else {
          for (int i2 = 0; i2 < v[i].size(); i2++) save.push_back(v[i][i2]);
        }
      }
      x = int(save.size()) % k;
      for (int i = 0; i < save.size() - x; i++) c[save[i]] = (i % k) + 1;
      for (int i = 0; i < n; i++) cout << c[i] << " ";
      cout << endl;
    }
    return 0;
  }
  