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 int INF = (is_same<int, long long>::value ? 1e18 + 666 : 1e9 + 666);
  mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
  template <class t1, class t2>
  inline bool cmin(t1 &a, const t2 &b) {
    if (a > b) {
      a = b;
      return true;
    }
    return false;
  }
  template <class t1, class t2>
  inline bool cmax(t1 &a, const t2 &b) {
    if (a < b) {
      a = b;
      return true;
    }
    return false;
  }
  inline int bpow(long long a, long long b, int mod) {
    int res = 1;
    a %= mod;
    for (; b; b >>= 1) {
      if (b & 1) {
        res = int((1ll * a * res) % mod);
        res %= mod;
      }
      a *= a;
      a %= mod;
    }
    return res;
  }
  void UseFiles(const string &s) {
    freopen((s + ".in").c_str(), "r", stdin);
    freopen((s + ".out").c_str(), "w", stdout);
  }
  void run();
  signed main() {
    iostream::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    run();
  }
  void solve();
  void run() {
    int n;
    cin >> n;
    while (n-- > 0) {
      solve();
    }
  }
  void solve() {
    int n, k;
    cin >> n >> k;
    map<int, vector<int>> pos;
    for (int i = 0; i < n; ++i) {
      int v;
      cin >> v;
      pos[v].push_back(i);
    }
    vector<int> positions;
    for (auto &[v, p] : pos) {
      while (((int)(p).size()) > k) {
        p.pop_back();
      }
      for (auto &x : p) {
        positions.push_back(x);
      }
    }
    vector<int> clr(n, -1);
    for (int i = 0; i + k <= ((int)(positions).size()); i += k) {
      for (int j = 0; j < k; ++j) {
        clr[positions[i + j]] = j;
      }
    }
    for (auto &c : clr) {
      cout << c + 1 << " ";
    }
    cout << endl;
  }
  