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;
  template <class T, class U>
  inline bool chmax(T& a, U b) {
    if (a < b) {
      a = b;
      return true;
    }
    return false;
  }
  template <class T, class U>
  inline bool chmin(T& a, U b) {
    if (a > b) {
      a = b;
      return true;
    }
    return false;
  }
  constexpr int INF = 1000000000;
  constexpr ll llINF = 3000000000000000000;
  constexpr int mod = 1000000007;
  constexpr double eps = 1e-10;
  const double pi = acos(-1);
  int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};
  int Random(int mi, int ma) {
    random_device rnd;
    mt19937 mt(rnd());
    uniform_int_distribution<int> engine(mi, ma);
    return engine(mt);
  }
  bool prime(int a) {
    if (a == 1) return false;
    for (int i = 2; i * i <= a; i++) {
      if (a % i == 0) return false;
    }
    return true;
  }
  vector<int> primes;
  void init_prime(int n) {
    primes.push_back(2);
    for (int i = 3; i <= n; i += 2) {
      bool f = true;
      for (int j : primes) {
        if (j * j > i) break;
        if (i % j == 0) {
          f = false;
          break;
        }
      }
      if (f) primes.push_back(i);
    }
  }
  ll modpow(ll a, ll b) {
    ll res = 1;
    while (b) {
      if (b & 1) {
        res *= a;
        res %= mod;
      }
      a *= a;
      a %= mod;
      b >>= 1;
    }
    return res;
  }
  vector<ll> inv, fact, factinv;
  void init_fact(int n) {
    inv.resize(n + 1);
    fact.resize(n + 1);
    factinv.resize(n + 1);
    inv[0] = 0;
    inv[1] = 1;
    fact[0] = 1;
    factinv[0] = 1;
    for (ll i = 1; i <= n; i++) {
      if (i >= 2) inv[i] = mod - ((mod / i) * inv[mod % i] % mod);
      fact[i] = (fact[i - 1] * i) % mod;
      factinv[i] = factinv[i - 1] * inv[i] % mod;
    }
  }
  ll _inv(ll a, ll m = mod) {
    ll b = m, u = 1, v = 0;
    while (b) {
      ll t = a / b;
      a -= t * b;
      swap(a, b);
      u -= t * v;
      swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
  }
  ll comb(int a, int b) {
    if (a < b) return 0;
    if (a < 0) return 0;
    return fact[a] * factinv[a - b] % mod * factinv[b] % mod;
  }
  void solve() {
    int n, k;
    cin >> n >> k;
    vector<vector<int>> v(n);
    for (int i = 0; i < n; i++) {
      int a;
      cin >> a;
      a--;
      v[a].push_back(i);
    }
    vector<int> ans(n);
    int s = 0;
    for (int i = 0; i < n; i++) s += min(k, (int)v[i].size());
    s = s / k * k;
    int cur = 0;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < min(k, (int)v[i].size()); j++) {
        if (cur < s) {
          ans[v[i][j]] = (cur % k) + 1;
          cur++;
        }
      }
    }
    for (int i = 0; i < n; i++) cout << ans[i] << " ";
    cout << endl;
  }
  int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int t;
    cin >> t;
    while (t--) solve();
  }
  