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 ceildiv(long long one, long long two) {
    if (one % two == 0) {
      return one / two;
    } else {
      return one / two + 1;
    }
  }
  long long power(long long n, long long pow, long long m) {
    if (pow == 0) return 1;
    if (pow % 2 == 0) {
      long long x = power(n, pow / 2, m);
      return (x * x) % m;
    } else
      return (power(n, pow - 1, m) * n) % m;
  }
  long long gcd(long long a, long long b) {
    if (!b) return a;
    return gcd(b, a % b);
  }
  long long factorial(long long n, long long mod) {
    if (n > 1)
      return (n * factorial(n - 1, mod)) % mod;
    else
      return 1;
  }
  long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
  vector<long long> read(long long n) {
    vector<long long> a;
    for (long long i = 0; i < n; i++) {
      long long x;
      cin >> x;
      a.push_back(x);
    }
    return a;
  }
  vector<vector<pair<long long, long long>>> adj;
  void solve() {
    long long n, k;
    cin >> n >> k;
    vector<long long> a = read(n);
    vector<long long> rep = a;
    sort(a.begin(), a.end());
    vector<vector<long long>> colors(k);
    queue<long long> q;
    for (long long i = 0; i < k; i++) {
      q.push(i);
    }
    for (long long i = 0; i < n; i++) {
      long long ind = q.front();
      long long x = upper_bound(colors[ind].begin(), colors[ind].end(), a[i]) -
                    lower_bound(colors[ind].begin(), colors[ind].end(), a[i]);
      if (x == 0) {
        colors[ind].push_back(a[i]);
        q.pop();
        q.push(ind);
      } else {
        continue;
      }
    }
    long long len = 1000000000;
    for (long long i = 0; i < k; i++) {
      len = min(len, (long long)colors[i].size());
    }
    for (long long i = 0; i < k; i++) {
      while (colors[i].size() > len) {
        colors[i].pop_back();
      }
    }
    map<long long, vector<long long>> mp;
    for (long long i = 0; i < k; i++) {
      for (auto x : colors[i]) {
        mp[x].push_back(i + 1);
      }
    }
    for (long long i = 0; i < n; i++) {
      if (mp[rep[i]].size() == 0) {
        cout << 0 << " ";
      } else {
        cout << mp[rep[i]].back() << " ";
        mp[rep[i]].pop_back();
      }
    }
    cout << '\n';
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    ;
    long long t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  