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 factorial(long long n) {
    long long res = 1, i;
    for (i = 2; i <= n; i++) {
      res *= i;
      res = res % 1000000007;
    }
    return res;
  }
  vector<long long> digits(long long n) {
    vector<long long> v;
    while (n) {
      v.push_back(n % 10);
      n /= 10;
    }
    return v;
  }
  vector<long long> SieveOfEratosthenes(long long n) {
    std::vector<long long> v(n + 1, 0);
    vector<long long> vb(n + 1, 1);
    long long i, j;
    std::vector<long long> ans;
    for (i = 2; i <= n; i++) {
      if (vb[i] == 1) {
        v[i] = i;
        ans.push_back(i);
        for (j = i * i; j <= n; j += i) {
          if (vb[i] == 1) {
            v[j] = i;
          }
          vb[j] = 0;
        }
      }
    }
    return ans;
  }
  map<long long, long long> get_prime_exponent(long long n) {
    std::vector<long long> v = SieveOfEratosthenes(n);
    long long i;
    long long cur = v[n];
    long long count = 0;
    map<long long, long long> m;
    while (n > 1) {
      cur = v[n];
      count = 0;
      while (n > 1 && n % cur == 0) {
        count++;
        n /= cur;
      }
      m[cur] = count;
    }
    return m;
  }
  void solve() {
    long long i, j, k, t, n;
    cin >> n >> j;
    std::vector<long long> v;
    for (i = 0; i <= n - 1; i++) {
      cin >> t;
      v.push_back(t);
    }
    std::vector<long long> vv = v;
    unordered_map<long long, long long> m;
    std::vector<std::vector<long long> > aa(n + 1);
    sort(v.begin(), v.end());
    long long color = 0;
    for (i = 0; i <= n - 1; i++) {
      if (m[v[i]] < j) {
        if (color == j) {
          color = 0;
        }
        aa[v[i]].push_back(color);
        color++;
      } else {
        aa[v[i]].push_back(-1);
      }
      m[v[i]]++;
    }
    std::vector<long long> colors(n, 0);
    if (color < j) {
      for (i = 0; i <= color - 1; i++) colors[i] = 1;
    }
    for (i = 0; i <= n - 1; i++) {
      t = aa[vv[i]].back();
      aa[vv[i]].pop_back();
      if (t == -1) {
        cout << 0;
      } else if (colors[t] == 0) {
        cout << t + 1;
      } else if (colors[t]) {
        cout << 0;
        colors[t]--;
      } else
        cout << 0;
      cout << " ";
    }
    cout << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    long long n;
    cin >> n;
    while (n--) {
      solve();
    }
    return 0;
  }
  