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 double pi = 2 * acos(0.0);
  template <typename T>
  T Sqr(T x) {
    T n = x * x;
    return n;
  }
  template <typename T>
  T Pow(T B, T P) {
    if (P == 0) return 1;
    if (P & 1)
      return B * Pow(B, P - 1);
    else
      return Sqr(Pow(B, P / 2));
  }
  template <typename T>
  T Abs(T a) {
    if (a < 0)
      return -a;
    else
      return a;
  }
  template <typename T>
  T Gcd(T a, T b) {
    if (a < 0) return Gcd(-a, b);
    if (b < 0) return Gcd(a, -b);
    return (b == 0) ? a : Gcd(b, a % b);
  }
  template <typename T>
  T Lcm(T a, T b) {
    if (a < 0) return Lcm(-a, b);
    if (b < 0) return Lcm(a, -b);
    return a * (b / Gcd(a, b));
  }
  template <typename T>
  T Exgcd(T a, T b, T &x, T &y) {
    if (a < 0) {
      T d = Exgcd(-a, b, x, y);
      x = -x;
      return d;
    }
    if (b < 0) {
      T d = Exgcd(a, -b, x, y);
      y = -y;
      return d;
    }
    if (b == 0) {
      x = 1;
      y = 0;
      return a;
    } else {
      T d = Exgcd(b, a % b, x, y);
      T t = x;
      x = y;
      y = t - (a / b) * y;
      return d;
    }
  }
  template <typename T>
  T BigMod(T b, T p, T m) {
    if (p == 0) return 1;
    if (p % 2 == 0) {
      T s = BigMod(b, p / 2, m);
      return ((s % m) * (s % m)) % m;
    }
    return ((b % m) * (BigMod(b, p - 1, m) % m)) % m;
  }
  template <typename T>
  T ModInvPrime(T b, T m) {
    return BigMod(b, m - 2, m);
  }
  template <typename T>
  T ModInvNotPrime(T a, T m) {
    T x, y;
    Exgcd(a, m, x, y);
    x %= m;
    if (x < 0) x += m;
    return x;
  }
  template <typename T>
  typename std::vector<T>::iterator Insert_sorted(std ::vector<T> &vec,
                                                  T const &item) {
    return vec.insert(std::upper_bound(vec.begin(), vec.end(), item), item);
  }
  template <typename T>
  inline string ToBinary(T n) {
    string r;
    while (n != 0) {
      r = (n % 2 == 0 ? "0" : "1") + r;
      n >>= 1;
    }
    return r;
  }
  long long BinaryToDecimal(string s) {
    int len = s.size();
    long long n = 0, p = 1;
    for (int i = len - 1; i >= 0; i--, p *= 2) n += p * (s[i] - '0');
    return n;
  }
  char Uplowch(char ch) {
    if (ch >= 'A' && ch <= 'Z') ch += 32;
    return ch;
  }
  char Lowupch(char ch) {
    if (ch >= 'a' && ch <= 'z') ch -= 32;
    return ch;
  }
  bool Isalpha(char ch) {
    if ((ch >= 'A' && ch <= 'Z') || (ch >= 'a' && ch <= 'z')) return true;
    return false;
  }
  int Strtoint(string str) {
    stringstream ss(str);
    int x = 0;
    ss >> x;
    return x;
  }
  string Intostr(int x) {
    stringstream ss;
    ss << x;
    string str = ss.str();
    return str;
  }
  vector<string> Linetostr(string str) {
    string s;
    vector<string> v;
    istringstream is(str);
    while (is >> s) v.push_back(s);
    return v;
  }
  template <typename T>
  void Print(T ar[], int a, int b) {
    for (int i = a; i + 1 <= b; i++) cout << ar[i] << " ";
    cout << ar[b] << "\n";
  }
  template <typename T>
  void Print(T ar[], int n) {
    for (int i = 0; i + 1 < n; i++) cout << ar[i] << " ";
    cout << ar[n - 1] << "\n";
  }
  template <typename T>
  void Print(const vector<T> &v) {
    for (int i = 0; i + 1 < v.size(); i++) cout << v[i] << " ";
    cout << v.back() << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int test;
    cin >> test;
    for (int tc = 1; tc <= test; tc++) {
      int n, k;
      cin >> n >> k;
      int ar[n + 1];
      for (int i = 0; i < n; i++) cin >> ar[i];
      int mp[n + 2];
      memset(mp, 0, sizeof(mp));
      vector<int> vec[n + 1];
      for (int i = 0; i < n; i++) {
        mp[ar[i]] += 1;
        vec[ar[i]].push_back(i);
      }
      int ans[n + 12];
      memset(ans, 0, sizeof(ans));
      vector<int> dewajabe;
      for (int i = 1; i <= n; i++) {
        if (mp[i] >= k) {
          for (int j = 0; j < k; j++) {
            ans[vec[i][j]] = j + 1;
          }
        } else if (mp[i] != 0) {
          for (int j = 0; j < vec[i].size(); j++) {
            dewajabe.push_back(vec[i][j]);
          }
        }
      }
      int now = dewajabe.size();
      for (int i = 0; i + k <= now;) {
        for (int j = 1; j <= k; j++, i++) {
          ans[dewajabe[i]] = j;
        }
      }
      for (int i = 0; i < n; i++) {
        cout << ans[i] << " ";
      }
      cout << "\n";
    }
    return 0;
  }
  