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;
  using ll = long long;
  using ull = unsigned long long;
  using ld = long double;
  using pii = pair<int, int>;
  using pll = pair<long long, long long>;
  const ll RANDOM =
      chrono::high_resolution_clock::now().time_since_epoch().count();
  const int INF = 0x3f3f3f3f;
  const ll LINF = 0x3f3f3f3f3f3f3f3f;
  const ll mod = 1e9 + 7;
  const int N = 2e5 + 5;
  int n, k;
  pii a[N];
  int d[N], ans[N];
  int c[N];
  void file() {
    {
      ios_base::sync_with_stdio(0);
      cout.tie(0);
      cin.tie(0);
    };
    if (fopen(""
              ".inp",
              "r")) {
      freopen(
          ""
          ".inp",
          "r", stdin);
      freopen(
          ""
          ".out",
          "w", stdout);
    }
  }
  void solve() {
    cin >> n >> k;
    memset(d, 0, sizeof d);
    for (int i = 1; i <= n; ++i)
      cin >> a[i].first, a[i].second = i, d[a[i].first]++;
    int num = 0;
    for (int i = 1; i <= n; ++i) {
      if (d[i] > k)
        num += k;
      else
        num += d[i];
    }
    num /= k;
    unordered_map<int, int> um[n + 1];
    int cnt = 0;
    sort(a + 1, a + n + 1);
    memset(ans, 0, sizeof ans);
    int color = 1;
    for (int i = 1; i <= n; ++i) {
      if (a[i].first == a[i - 1].first) {
        if (!um[a[i].first].count(color)) {
          ans[a[i].second] = color;
          um[a[i].first][color++] = 1;
          if (color > k) {
            if (++cnt >= num) break;
            color = 1;
          }
        } else
          ans[a[i].second] = 0;
      } else {
        ans[a[i].second] = color;
        um[a[i].first][color++] = 1;
        if (color > k) {
          if (++cnt >= num) break;
          color = 1;
        }
      }
    }
    for (int i = 1; i <= n; ++i) cout << ans[i] << ' ';
    cout << '\n';
  }
  int main() {
    file();
    int t;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  