#include <bits/stdc++.h>
using namespace std;
char local = 'O';
template <typename T>
void read(T &p) {
  cin >> p;
}
template <typename T, typename T1>
void read(pair<T, T1> &p) {
  read(p.first);
  read(p.second);
}
template <typename T>
void read(T arr[], long long n) {
  for (long long i = 0; i < n; i++) {
    read(arr[i]);
  }
}
template <typename T>
void read(vector<T> &arr) {
  for (long long i = 0; i < arr.size(); i++) {
    read(arr[i]);
  }
}
template <typename T>
void out(T &n) {
  cout << n;
}
template <typename T, typename T1>
void out(pair<T, T1> &p) {
  cout << "(";
  out(p.first);
  cout << ",";
  out(p.second);
  cout << ")";
}
template <typename T>
void out(T arr[], long long n) {
  for (long long i = 0; i < n; i++) {
    out(arr[i]);
    cout << " ";
  }
}
template <typename T>
void out(vector<T> &arr) {
  cout << "[ ";
  for (long long i = 0; i < arr.size(); i++) {
    out(arr[i]);
    if (i != arr.size() - 1) cout << ", ";
  }
  cout << " ]";
}
template <typename T>
void out(set<T> &arr) {
  for (auto i : arr) {
    out(i);
    cout << " ";
  }
}
template <typename T>
void out(multiset<T> &arr) {
  for (auto i : arr) {
    out(i);
    cout << " ";
  }
}
template <typename T, typename T1>
void out(map<T, T1> &m) {
  for (auto i : m) {
    out(i.first);
    cout << " -> ";
    out(i.second);
    cout << ", ";
  }
}
template <typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
  cout << name << " : ";
  out(arg1);
}
template <typename Arg1, typename... Args>
void __f(const char *names, Arg1 &&arg1, Args &&...args) {
  const char *comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : ";
  out(arg1);
  cout << " | ";
  __f(comma + 1, args...);
}
void cal() {}
void solve() {
  int n, k;
  cin >> n >> k;
  vector<int> v(n);
  read(v);
  int c = 0;
  map<int, vector<int>> m;
  for (int i = 0; i < n; i++) {
    m[v[i]].push_back(i);
  }
  vector<int> res(n, 0);
  int make_pair = 0;
  for (auto i : m) {
    int ic = k;
    for (auto x : i.second) {
      if (ic == 0) {
        break;
      }
      res[x] = c + 1;
      c++;
      make_pair += c / k;
      c %= k;
      ic--;
    }
  }
  map<int, int> mc;
  for (auto &i : res) {
    if (mc[i] == make_pair) {
      i = 0;
    } else {
      mc[i]++;
    }
    cout << i << " ";
  }
  cout << "\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t1;
  t1 = 1;
  long long k = 1;
  cin >> t1;
  cal();
  while (t1--) {
    solve();
  }
}

### Prompt

Your task is to create a cpp solution 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;
char local = 'O';
template <typename T>
void read(T &p) {
  cin >> p;
}
template <typename T, typename T1>
void read(pair<T, T1> &p) {
  read(p.first);
  read(p.second);
}
template <typename T>
void read(T arr[], long long n) {
  for (long long i = 0; i < n; i++) {
    read(arr[i]);
  }
}
template <typename T>
void read(vector<T> &arr) {
  for (long long i = 0; i < arr.size(); i++) {
    read(arr[i]);
  }
}
template <typename T>
void out(T &n) {
  cout << n;
}
template <typename T, typename T1>
void out(pair<T, T1> &p) {
  cout << "(";
  out(p.first);
  cout << ",";
  out(p.second);
  cout << ")";
}
template <typename T>
void out(T arr[], long long n) {
  for (long long i = 0; i < n; i++) {
    out(arr[i]);
    cout << " ";
  }
}
template <typename T>
void out(vector<T> &arr) {
  cout << "[ ";
  for (long long i = 0; i < arr.size(); i++) {
    out(arr[i]);
    if (i != arr.size() - 1) cout << ", ";
  }
  cout << " ]";
}
template <typename T>
void out(set<T> &arr) {
  for (auto i : arr) {
    out(i);
    cout << " ";
  }
}
template <typename T>
void out(multiset<T> &arr) {
  for (auto i : arr) {
    out(i);
    cout << " ";
  }
}
template <typename T, typename T1>
void out(map<T, T1> &m) {
  for (auto i : m) {
    out(i.first);
    cout << " -> ";
    out(i.second);
    cout << ", ";
  }
}
template <typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
  cout << name << " : ";
  out(arg1);
}
template <typename Arg1, typename... Args>
void __f(const char *names, Arg1 &&arg1, Args &&...args) {
  const char *comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : ";
  out(arg1);
  cout << " | ";
  __f(comma + 1, args...);
}
void cal() {}
void solve() {
  int n, k;
  cin >> n >> k;
  vector<int> v(n);
  read(v);
  int c = 0;
  map<int, vector<int>> m;
  for (int i = 0; i < n; i++) {
    m[v[i]].push_back(i);
  }
  vector<int> res(n, 0);
  int make_pair = 0;
  for (auto i : m) {
    int ic = k;
    for (auto x : i.second) {
      if (ic == 0) {
        break;
      }
      res[x] = c + 1;
      c++;
      make_pair += c / k;
      c %= k;
      ic--;
    }
  }
  map<int, int> mc;
  for (auto &i : res) {
    if (mc[i] == make_pair) {
      i = 0;
    } else {
      mc[i]++;
    }
    cout << i << " ";
  }
  cout << "\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t1;
  t1 = 1;
  long long k = 1;
  cin >> t1;
  cal();
  while (t1--) {
    solve();
  }
}

```