#include <bits/stdc++.h>
using namespace std;
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    int arr[n];
    for (int i = 0; i < n; i++) {
      cin >> arr[i];
    }
    map<int, int> freq;
    for (int i = 0; i < n; i++) {
      freq[arr[i]] += 1;
    }
    vector<pair<int, int>> v;
    map<int, int>::iterator it;
    for (it = freq.begin(); it != freq.end(); it++) {
      v.push_back({it->second, it->first});
    }
    sort(v.begin(), v.end());
    int k_copy = k;
    map<int, queue<int>> ans;
    vector<int> remaining;
    for (int i = 0; i < v.size(); i++) {
      int freqs = v[i].first;
      int num = v[i].second;
      if (freqs >= k) {
        queue<int> s;
        for (int j = 1; j <= k; j++) {
          s.push(j);
        }
        ans[num] = s;
      } else {
        for (int j = 0; j < freqs; j++) {
          remaining.push_back(num);
        }
      }
    }
    vector<int> temp;
    for (int i = 0; i < remaining.size(); i++) {
      if (temp.size() < k) {
        temp.push_back(remaining[i]);
      } else {
        for (int j = 0; j < k; j++) {
          if (ans.find(temp[j]) == ans.end()) {
            queue<int> s;
            s.push(j + 1);
            ans[temp[j]] = s;
          } else {
            ans[temp[j]].push(j + 1);
          }
        }
        temp.clear();
        temp.push_back(remaining[i]);
      }
    }
    if (temp.size() == k) {
      for (int j = 0; j < k; j++) {
        if (ans.find(temp[j]) == ans.end()) {
          queue<int> s;
          s.push(j + 1);
          ans[temp[j]] = s;
        } else {
          ans[temp[j]].push(j + 1);
        }
      }
    }
    for (int i = 0; i < n; i++) {
      if (ans[arr[i]].empty()) {
        cout << 0 << " ";
      } else {
        cout << ans[arr[i]].front() << " ";
        ans[arr[i]].pop();
      }
    }
    cout << (char)(10);
  }
  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;
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    int arr[n];
    for (int i = 0; i < n; i++) {
      cin >> arr[i];
    }
    map<int, int> freq;
    for (int i = 0; i < n; i++) {
      freq[arr[i]] += 1;
    }
    vector<pair<int, int>> v;
    map<int, int>::iterator it;
    for (it = freq.begin(); it != freq.end(); it++) {
      v.push_back({it->second, it->first});
    }
    sort(v.begin(), v.end());
    int k_copy = k;
    map<int, queue<int>> ans;
    vector<int> remaining;
    for (int i = 0; i < v.size(); i++) {
      int freqs = v[i].first;
      int num = v[i].second;
      if (freqs >= k) {
        queue<int> s;
        for (int j = 1; j <= k; j++) {
          s.push(j);
        }
        ans[num] = s;
      } else {
        for (int j = 0; j < freqs; j++) {
          remaining.push_back(num);
        }
      }
    }
    vector<int> temp;
    for (int i = 0; i < remaining.size(); i++) {
      if (temp.size() < k) {
        temp.push_back(remaining[i]);
      } else {
        for (int j = 0; j < k; j++) {
          if (ans.find(temp[j]) == ans.end()) {
            queue<int> s;
            s.push(j + 1);
            ans[temp[j]] = s;
          } else {
            ans[temp[j]].push(j + 1);
          }
        }
        temp.clear();
        temp.push_back(remaining[i]);
      }
    }
    if (temp.size() == k) {
      for (int j = 0; j < k; j++) {
        if (ans.find(temp[j]) == ans.end()) {
          queue<int> s;
          s.push(j + 1);
          ans[temp[j]] = s;
        } else {
          ans[temp[j]].push(j + 1);
        }
      }
    }
    for (int i = 0; i < n; i++) {
      if (ans[arr[i]].empty()) {
        cout << 0 << " ";
      } else {
        cout << ans[arr[i]].front() << " ";
        ans[arr[i]].pop();
      }
    }
    cout << (char)(10);
  }
  return 0;
}

```