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 int maxn = 2e5 + 5;
  const int mod = 1e9 + 7;
  int t, n, k;
  int a[maxn], num[maxn];
  int NUM[maxn], vis[maxn];
  struct node {
    int val;
    int pos;
    int color;
  } it[maxn];
  struct nnode {
    int l, r;
  } pos[maxn];
  bool cmp1(node a, node b) {
    if (num[a.val] != num[b.val])
      return num[a.val] > num[b.val];
    else if (a.val != b.val)
      return a.val < b.val;
    else
      return a.pos < b.pos;
  }
  bool cmp2(node a, node b) { return a.pos < b.pos; }
  int main() {
    cin >> t;
    while (t--) {
      cin >> n >> k;
      int ans = 0;
      memset(NUM, 0, sizeof(NUM));
      memset(vis, 0, sizeof(vis));
      memset(pos, 0, sizeof(pos));
      int nnum = 0;
      for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        it[i].val = a[i];
        it[i].pos = i;
        it[i].color = 0;
        num[i] = 0;
      }
      for (int i = 1; i <= n; i++) num[a[i]]++;
      for (int i = 1; i <= n; i++) {
        if (num[i]) {
          NUM[nnum++] = i;
          ans += min(num[i], k);
        }
      }
      int cnt = ans / k;
      sort(it + 1, it + 1 + n, cmp1);
      int all = cnt * k;
      int last = 0;
      for (int i = 1; i <= n; i++) {
        int nn = num[it[i].val], nnn = it[i].val;
        if (nn >= k) {
          int col = 1, j;
          for (j = i; j < i + k; j++) it[j].color = col++;
          while (it[j].val == nnn) j++;
          i = j - 1;
          all -= k;
        } else {
          last = i;
          break;
        }
      }
      for (int i = last; i <= n; i++) {
        if (all == 0) break;
        if ((i - last + 1) % k == 0)
          it[i].color = k;
        else
          it[i].color = (i - last + 1) % k;
        all--;
      }
      sort(it + 1, it + 1 + n, cmp2);
      for (int i = 1; i <= n; i++) {
        printf("%d ", it[i].color);
      }
      printf("\n");
    }
    return 0;
  }
  