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>
  #pragma GCC optimize("Ofast")
  #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
  #pragma GCC optimize("unroll-loops")
  using namespace std;
  long long MOD = 998244353;
  long long mod = 1000000007;
  double eps = 1e-12;
  double pi = acos(-1);
  mt19937 RNG(chrono::steady_clock::now().time_since_epoch().count());
  template <typename T>
  T gcd(T a, T b) {
    return (b ? __gcd(a, b) : a);
  }
  template <typename T>
  T lcm(T a, T b) {
    return (a * (b / gcd(a, b)));
  }
  int add(int a, int b, int c = MOD) {
    int res = a + b;
    return (res >= c ? res - c : res);
  }
  int mod_neg(int a, int b, int c = MOD) {
    int res;
    if (abs(a - b) < c)
      res = a - b;
    else
      res = (a - b) % c;
    return (res < 0 ? res + c : res);
  }
  int mul(int a, int b, int c = MOD) {
    long long res = (long long)a * b;
    return (res >= c ? res % c : res);
  }
  int muln(int a, int b, int c = MOD) {
    long long res = (long long)a * b;
    return ((res % c) + c) % c;
  }
  long long mulmod(long long a, long long b, long long m = MOD) {
    long long q = (long long)(((long double)a * (long double)b) / (long double)m);
    long long r = a * b - q * m;
    if (r > m) r %= m;
    if (r < 0) r += m;
    return r;
  }
  template <typename T>
  T expo(T e, T n) {
    T x = 1, p = e;
    while (n) {
      if (n & 1) x = x * p;
      p = p * p;
      n >>= 1;
    }
    return x;
  }
  template <typename T>
  T power(T e, T n, T m = MOD) {
    T x = 1, p = e;
    while (n) {
      if (n & 1) x = mul(x, p, m);
      p = mul(p, p, m);
      n >>= 1;
    }
    return x;
  }
  template <typename T>
  T extended_euclid(T a, T b, T &x, T &y) {
    T xx = 0, yy = 1;
    y = 0;
    x = 1;
    while (b) {
      T q = a / b, t = b;
      b = a % b;
      a = t;
      t = xx;
      xx = x - q * xx;
      x = t;
      t = yy;
      yy = y - q * yy;
      y = t;
    }
    return a;
  }
  template <typename T>
  T mod_inverse(T a, T n = MOD) {
    T x, y, z = 0;
    T d = extended_euclid(a, n, x, y);
    return (d > 1 ? -1 : mod_neg(x, z, n));
  }
  const int FACSZ = 1;
  int fact[FACSZ], ifact[FACSZ];
  void precom(int c = MOD) {
    fact[0] = 1;
    for (long long i = 1; i < FACSZ; i++) fact[i] = mul(fact[i - 1], i, c);
    ifact[FACSZ - 1] = mod_inverse(fact[FACSZ - 1], c);
    for (long long i = 0; i < FACSZ - 1; i++) {
      ifact[i] = mul(i + 1, ifact[i + 1], c);
    }
  }
  int ncr(int n, int r, int c = MOD) {
    return mul(mul(ifact[r], ifact[n - r], c), fact[n], c);
  }
  void solve() {
    long long n, k;
    cin >> n >> k;
    long long a[n];
    map<int, vector<int> > m;
    vector<long long> ans(n, 0);
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
      m[a[i]].push_back(i);
    }
    long long count = 0;
    long long count2 = 0;
    for (auto x : m) {
      if (x.second.size() < k) {
        count += x.second.size();
      } else {
        count2++;
        for (long long i = 0; i < k; i++) {
          ans[x.second[i]] = i + 1;
        }
      }
    }
    long long g = 0;
    for (auto x : m) {
      if (x.second.size() < k) {
        for (long long i = 0; i < x.second.size(); i++) {
          if (g < count - count % k) {
            ans[x.second[i]] = g % k + 1;
            g++;
          }
        }
      }
    }
    for (long long i = 0; i < n; i++) {
      cout << ans[i] << " ";
    }
    cout << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long t;
    cin >> t;
    for (int it = 1; it <= t; it++) {
      solve();
    }
    return 0;
  }
  