#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a / gcd(a, b) * b); }
bool sorta(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
bool sortd(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second > b.second);
}
void printarr(long long arr[], long long n) {
  for (int i = 0; i < n; i++) cout << arr[i] << " ";
  cout << "\n";
}
void asquare() {
  long long n, k, nn = 0, mm = 0, tt, x;
  cin >> n >> k;
  long long arr[n];
  long long ans[n];
  for (int i = 0; i < n; i++) cin >> arr[i];
  unordered_map<long long, vector<long long>> ump;
  unordered_map<long long, long long> mmm;
  unordered_map<long long, vector<long long>> uv;
  vector<long long> v;
  for (int i = 0; i < n; i++) {
    ump[arr[i]].push_back(i);
  }
  long long index = 1, store;
  for (auto &a : ump) {
    if (a.second.size() >= k) {
      mm++;
      for (int i = 0; i < k; i++) {
        ans[a.second[i]] = index;
        index++;
        if (index == k + 1) index = 1;
      }
      for (int i = 0; i < a.second.size() - k; i++) ans[a.second[i + k]] = 0;
    } else {
      store = index;
      for (int i = 0; i < a.second.size(); i++) {
        ans[a.second[i]] = index;
        index++;
        if (index == k + 1) index = 1;
      }
      nn += a.second.size();
    }
  }
  tt = mm + nn / k;
  for (int i = 0; i < n; i++) {
    mmm[ans[i]]++;
    if (mmm[ans[i]] <= tt)
      cout << ans[i] << " ";
    else
      cout << "0 ";
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    asquare();
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a / gcd(a, b) * b); }
bool sorta(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
bool sortd(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second > b.second);
}
void printarr(long long arr[], long long n) {
  for (int i = 0; i < n; i++) cout << arr[i] << " ";
  cout << "\n";
}
void asquare() {
  long long n, k, nn = 0, mm = 0, tt, x;
  cin >> n >> k;
  long long arr[n];
  long long ans[n];
  for (int i = 0; i < n; i++) cin >> arr[i];
  unordered_map<long long, vector<long long>> ump;
  unordered_map<long long, long long> mmm;
  unordered_map<long long, vector<long long>> uv;
  vector<long long> v;
  for (int i = 0; i < n; i++) {
    ump[arr[i]].push_back(i);
  }
  long long index = 1, store;
  for (auto &a : ump) {
    if (a.second.size() >= k) {
      mm++;
      for (int i = 0; i < k; i++) {
        ans[a.second[i]] = index;
        index++;
        if (index == k + 1) index = 1;
      }
      for (int i = 0; i < a.second.size() - k; i++) ans[a.second[i + k]] = 0;
    } else {
      store = index;
      for (int i = 0; i < a.second.size(); i++) {
        ans[a.second[i]] = index;
        index++;
        if (index == k + 1) index = 1;
      }
      nn += a.second.size();
    }
  }
  tt = mm + nn / k;
  for (int i = 0; i < n; i++) {
    mmm[ans[i]]++;
    if (mmm[ans[i]] <= tt)
      cout << ans[i] << " ";
    else
      cout << "0 ";
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    asquare();
  }
  return 0;
}

```