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;
  using ll = long long;
  using ld = long double;
  using ull = unsigned long long;
  const int MOD = 1e9 + 7;
  const ll INF = 1e18;
  inline void fastio() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
  }
  template <typename A, typename B>
  ostream &operator<<(ostream &os, const pair<A, B> &p) {
    return os << '(' << p.first << ", " << p.second << ')';
  }
  template <typename T_container, typename T = typename enable_if<
                                      !is_same<T_container, string>::value,
                                      typename T_container::value_type>::type>
  ostream &operator<<(ostream &os, const T_container &v) {
    os << '{';
    string sep;
    for (const T &x : v) os << sep << x, sep = ", ";
    return os << '}';
  }
  void dbg_out() { cerr << endl; }
  template <typename Head, typename... Tail>
  void dbg_out(Head H, Tail... T) {
    cerr << ' ' << H;
    dbg_out(T...);
  }
  mt19937 rng(
      (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
  int rand(int l, int r) {
    uniform_int_distribution<int> uid(l, r);
    return uid(rng);
  }
  void solve() {
    int n, k;
    cin >> n >> k;
    vector<int> arr(n);
    map<int, vector<int>> ids;
    for (int i = 0; i < n; i++) {
      cin >> arr[i];
      ids[arr[i]].push_back(i);
    }
    vector<int> cols[k];
    vector<int> out(n);
    int cur{};
    for (auto &p : ids) {
      for (int i = 0; i < min((int)(p.second).size(), k); i++) {
        cols[cur].push_back(p.second[i]);
        cur = (cur + 1) % k;
      }
    }
    int mn = INT_MAX;
    for (auto &v : cols) mn = min(mn, (int)(v).size());
    for (int col = 0; col < k; col++) {
      if ((int)(cols[col]).size() > mn) {
        cols[col].pop_back();
      }
      for (int idx : cols[col]) {
        out[idx] = col + 1;
      }
    }
    for (int e : out) {
      cout << e << " ";
    }
    cout << '\n';
  }
  int main() {
    fastio();
    cout << setprecision(15) << fixed;
    int tc = 1;
    cin >> tc;
    while (tc--) {
      solve();
    }
    return 0;
  }
  