#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using ull = unsigned long long;
const int MOD = 1e9 + 7;
const ll INF = 1e18;
inline void fastio() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
template <typename T_container, typename T = typename enable_if<
                                    !is_same<T_container, string>::value,
                                    typename T_container::value_type>::type>
ostream &operator<<(ostream &os, const T_container &v) {
  os << '{';
  string sep;
  for (const T &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
void dbg_out() { cerr << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << ' ' << H;
  dbg_out(T...);
}
mt19937 rng(
    (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
int rand(int l, int r) {
  uniform_int_distribution<int> uid(l, r);
  return uid(rng);
}
void solve() {
  int n, k;
  cin >> n >> k;
  vector<int> arr(n);
  map<int, vector<int>> ids;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    ids[arr[i]].push_back(i);
  }
  vector<int> cols[k];
  vector<int> out(n);
  int cur{};
  for (auto &p : ids) {
    for (int i = 0; i < min((int)(p.second).size(), k); i++) {
      cols[cur].push_back(p.second[i]);
      cur = (cur + 1) % k;
    }
  }
  int mn = INT_MAX;
  for (auto &v : cols) mn = min(mn, (int)(v).size());
  for (int col = 0; col < k; col++) {
    if ((int)(cols[col]).size() > mn) {
      cols[col].pop_back();
    }
    for (int idx : cols[col]) {
      out[idx] = col + 1;
    }
  }
  for (int e : out) {
    cout << e << " ";
  }
  cout << '\n';
}
int main() {
  fastio();
  cout << setprecision(15) << fixed;
  int tc = 1;
  cin >> tc;
  while (tc--) {
    solve();
  }
  return 0;
}

### Prompt

Please create a solution in CPP to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using ull = unsigned long long;
const int MOD = 1e9 + 7;
const ll INF = 1e18;
inline void fastio() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
template <typename T_container, typename T = typename enable_if<
                                    !is_same<T_container, string>::value,
                                    typename T_container::value_type>::type>
ostream &operator<<(ostream &os, const T_container &v) {
  os << '{';
  string sep;
  for (const T &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
void dbg_out() { cerr << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << ' ' << H;
  dbg_out(T...);
}
mt19937 rng(
    (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
int rand(int l, int r) {
  uniform_int_distribution<int> uid(l, r);
  return uid(rng);
}
void solve() {
  int n, k;
  cin >> n >> k;
  vector<int> arr(n);
  map<int, vector<int>> ids;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    ids[arr[i]].push_back(i);
  }
  vector<int> cols[k];
  vector<int> out(n);
  int cur{};
  for (auto &p : ids) {
    for (int i = 0; i < min((int)(p.second).size(), k); i++) {
      cols[cur].push_back(p.second[i]);
      cur = (cur + 1) % k;
    }
  }
  int mn = INT_MAX;
  for (auto &v : cols) mn = min(mn, (int)(v).size());
  for (int col = 0; col < k; col++) {
    if ((int)(cols[col]).size() > mn) {
      cols[col].pop_back();
    }
    for (int idx : cols[col]) {
      out[idx] = col + 1;
    }
  }
  for (int e : out) {
    cout << e << " ";
  }
  cout << '\n';
}
int main() {
  fastio();
  cout << setprecision(15) << fixed;
  int tc = 1;
  cin >> tc;
  while (tc--) {
    solve();
  }
  return 0;
}

```