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 T1, typename T2>
  inline ostream &operator<<(ostream &os, const pair<T1, T2> &p) {
    return os << "(" << p.first << ", " << p.second << ")";
  }
  template <typename T, size_t N>
  inline ostream &operator<<(ostream &os, const array<T, N> &a) {
    os << "[";
    int cnt = 0;
    for (auto &val : a) {
      if (cnt++) os << ", ";
      os << val;
    }
    os << "]";
    return os;
  }
  template <typename T>
  inline ostream &operator<<(ostream &os, const vector<T> &v) {
    os << "[";
    int cnt = 0;
    for (auto &val : v) {
      if (cnt++) os << ", ";
      os << val;
    }
    return os << "]";
  }
  template <typename T>
  inline ostream &operator<<(ostream &os, const set<T> &v) {
    os << "{";
    int cnt = 0;
    for (auto &val : v) {
      if (cnt++) os << ", ";
      os << val;
    }
    return os << "}";
  }
  template <typename T>
  inline ostream &operator<<(ostream &os, const unordered_set<T> &v) {
    os << "{";
    int cnt = 0;
    for (auto &val : v) {
      if (cnt++) os << ", ";
      os << val;
    }
    return os << "}";
  }
  template <typename T>
  inline ostream &operator<<(ostream &os, const multiset<T> &v) {
    os << "{";
    int cnt = 0;
    for (auto &val : v) {
      if (cnt++) os << ", ";
      os << val;
    }
    return os << "}";
  }
  template <typename T1, typename T2>
  inline ostream &operator<<(ostream &os, const map<T1, T2> &v) {
    os << "[";
    int cnt = 0;
    for (auto &val : v) {
      if (cnt++) os << ", ";
      os << val;
    }
    return os << "]";
  }
  template <typename Arg1>
  void __f(const char *name, Arg1 &&arg1) {
    cerr << name << ": " << arg1 << endl;
  }
  template <typename Arg1, typename... Args>
  void __f(const char *names, Arg1 &&arg1, Args &&...args) {
    const char *comma = strchr(names + 1, ',');
    cerr.write(names, comma - names) << ": " << arg1 << " |";
    __f(comma + 1, args...);
  }
  using ll = long long;
  using ull = unsigned long long;
  using vi = vector<int>;
  using vll = vector<long long>;
  using vvi = vector<vector<int>>;
  using vs = vector<string>;
  using vb = vector<bool>;
  using vd = vector<double>;
  using vvb = vector<vector<bool>>;
  using vvc = vector<vector<char>>;
  using vc = vector<char>;
  using si = set<int>;
  using mpii = map<int, int>;
  using pii = pair<int, int>;
  using pll = pair<ll, ll>;
  const int dx[] = {-1, 1, 0, 0};
  const int dy[] = {0, 0, 1, -1};
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  template <class T>
  inline bool umin(T &a, const T b) {
    return b < a ? a = b, 1 : 0;
  }
  template <class T>
  inline bool umax(T &a, const T b) {
    return a < b ? a = b, 1 : 0;
  }
  void solve() {
    int n, k;
    cin >> n >> k;
    vi a(n);
    map<int, vi> pos;
    int rem = 0;
    for (int i = 0; i < (n); i++) {
      cin >> a[i];
      if (((int)(pos[a[i]]).size()) < k) {
        pos[a[i]].push_back(i);
        rem++;
      }
    }
    vi keys;
    for (auto &[c, ps] : pos) {
      keys.emplace_back(c);
    }
    vi b(n, 0);
    int m = ((int)(keys).size());
    int i = 0;
    while (rem >= k) {
      for (int j = 0; j < (k); j++) {
        while (i < m && ((int)(pos[keys[i]]).size()) == 0) i++;
        if (i == m) continue;
        int p = pos[keys[i]].back();
        pos[keys[i]].pop_back();
        b[p] = j + 1;
      }
      rem -= k;
    }
    for (int j = 0; j < (n); j++) cout << b[j] << " \n"[j == n - 1];
  }
  void solve(int _cas) { solve(); }
  int main() {
    srand(time(NULL));
    ios_base::sync_with_stdio(false);
    cin.tie(NULL), cout.tie(NULL), cout.precision(12);
    int cas;
    cin >> cas;
    for (int i = 0; i < (cas); i++) solve(i + 1);
  }
  