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 long long mod = 1e9 + 7;
  const long long mxn = 1e5 + 7;
  const long long inf = 1e18;
  inline void normal(long long &a) {
    a %= mod;
    (a < 0) && (a += mod);
  }
  inline long long modMul(long long a, long long b) {
    a %= mod, b %= mod;
    normal(a), normal(b);
    return (a * b) % mod;
  }
  inline long long modAdd(long long a, long long b) {
    a %= mod, b %= mod;
    normal(a), normal(b);
    return (a + b) % mod;
  }
  inline long long modSub(long long a, long long b) {
    a %= mod, b %= mod;
    normal(a), normal(b);
    a -= b;
    normal(a);
    return a;
  }
  inline long long modPow(long long b, long long p) {
    long long r = 1;
    while (p) {
      if (p & 1) r = modMul(r, b);
      b = modMul(b, b);
      p >>= 1;
    }
    return r;
  }
  inline long long modInverse(long long a) { return modPow(a, mod - 2); }
  inline long long modDiv(long long a, long long b) {
    return modMul(a, modInverse(b));
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);
    cout.tie(NULL);
    ;
    long long i, j, k, a, b, c, d, n, m, t, u, v, w, x, y, z;
    cin >> t;
    for (long long zz = 1; zz <= t; zz++) {
      cin >> n >> k;
      vector<long long> vec(n);
      set<long long> st;
      map<long long, long long> frq;
      map<long long, vector<long long> > pos;
      for (i = 0; i < n; i++) {
        cin >> vec[i];
        a = vec[i];
        st.insert(a);
        frq[a] += 1;
        pos[a].push_back(i);
      }
      vector<long long> ans(n, 0), tem1, tem2;
      for (auto it : st) {
        if (frq[it] >= k) {
          tem1.push_back(it);
        } else {
          tem2.push_back(it);
        }
      }
      for (auto a : tem1) {
        long long col = 0;
        for (auto b : pos[a]) {
          col += 1;
          ans[b] = col;
          if (col == k) {
            break;
          }
        }
      }
      long long col = 0, cnt = (long long)tem1.size();
      for (auto a : tem2) {
        for (auto b : pos[a]) {
          long long tem = (col % k) + 1;
          ans[b] = tem;
          if (tem == k) {
            cnt += 1;
          }
          col += 1;
        }
      }
      map<long long, long long> check;
      for (i = 0; i < n; i++) {
        a = ans[i];
        if (a >= 1) {
          check[a] += 1;
        }
        if (check[a] <= cnt) {
          cout << ans[i] << " ";
        } else {
          cout << 0 << " ";
        }
      }
      cout << "\n";
    }
    return 0;
  }
  