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 power(long long x, long long y) {
    long long res = 1;
    while (y > 0) {
      if (y & 1) res = res * x;
      y = y >> 1;
      x = x * x;
    }
    return res;
  }
  long long gcd(long long a, long long b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long pow(long long a, long long b, long long MOD) {
    long long x = 1, y = a;
    while (b > 0) {
      if (b % 2 == 1) {
        x = (x % MOD * y % MOD);
        if (x > MOD) x %= MOD;
      }
      y = (y % MOD * y % MOD);
      if (y > MOD) y %= MOD;
      b /= 2;
    }
    return x;
  }
  void union1(long long parent[], int x, int y) { parent[y] = x; }
  int find1(long long parent[], int x) {
    if (parent[x] == -1) {
      return x;
    }
    return find1(parent, parent[x]);
  }
  int main(void) {
    long long t, n, x, y, k, m, r, p, d, u, v, q, a, b, z, w, l;
    cin >> t;
    while (t--) {
      cin >> n >> k;
      unordered_map<long long, long long> mp;
      unordered_map<long long, vector<long long>> v1;
      unordered_map<long long, long long> indices;
      vector<long long> v(n);
      for (int i = 0; i < n; i++) {
        cin >> v[i];
        mp[v[i]]++;
      }
      vector<pair<int, int>> p;
      long long res = 0;
      for (auto x : mp) {
        if (x.second >= k) {
          long long r = k, c = 1;
          while (r--) {
            v1[x.first].push_back(c++);
          }
        } else {
          p.push_back({x.second, x.first});
          res += x.second;
        }
      }
      int f = 0, c1 = 0;
      long long temp = res % k;
      res -= temp;
      for (int i = 0; i < p.size(); i++) {
        while (c1 < res && p[i].first--) {
          c1++;
          v1[p[i].second].push_back((f % k) + 1);
          f++;
        }
      }
      for (auto x : v1) {
        indices[x.first] = 0;
      }
      for (int i = 0; i < n; i++) {
        if (indices[v[i]] < v1[v[i]].size()) {
          cout << v1[v[i]][indices[v[i]]++] << " ";
        } else
          cout << 0 << " ";
      }
      cout << endl;
    }
  }
  