#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);
  }
};
void solve() {
  int n, numColors;
  cin >> n >> numColors;
  int k = numColors;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  vector<int> color(n, 0);
  unordered_map<int, vector<int>> colormap;
  for (int i = 0; i < n; i++) {
    if (colormap.find(a[i]) == colormap.end()) {
      vector<int> v;
      v.push_back(i);
      colormap[a[i]] = v;
    } else {
      colormap[a[i]].push_back(i);
    }
  }
  vector<int> multicolor;
  for (auto itr = colormap.begin(); itr != colormap.end(); itr++) {
    if (itr->second.size() >= k) {
      for (int i = 0; i < k; i++) {
        color[itr->second[i]] = i + 1;
      }
    } else {
      for (int i = 0; i < itr->second.size(); i++) {
        multicolor.push_back(itr->second[i]);
      }
    }
  }
  int mc = multicolor.size();
  int counter = 0;
  int stoper = mc - mc % k;
  for (int i : multicolor) {
    if (counter == stoper) break;
    color[i] = (counter % k) + 1;
    counter++;
  }
  for (int ans : color) cout << ans << " ";
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tc;
  cin >> tc;
  for (int t = 1; t <= tc; t++) {
    solve();
  }
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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);
  }
};
void solve() {
  int n, numColors;
  cin >> n >> numColors;
  int k = numColors;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  vector<int> color(n, 0);
  unordered_map<int, vector<int>> colormap;
  for (int i = 0; i < n; i++) {
    if (colormap.find(a[i]) == colormap.end()) {
      vector<int> v;
      v.push_back(i);
      colormap[a[i]] = v;
    } else {
      colormap[a[i]].push_back(i);
    }
  }
  vector<int> multicolor;
  for (auto itr = colormap.begin(); itr != colormap.end(); itr++) {
    if (itr->second.size() >= k) {
      for (int i = 0; i < k; i++) {
        color[itr->second[i]] = i + 1;
      }
    } else {
      for (int i = 0; i < itr->second.size(); i++) {
        multicolor.push_back(itr->second[i]);
      }
    }
  }
  int mc = multicolor.size();
  int counter = 0;
  int stoper = mc - mc % k;
  for (int i : multicolor) {
    if (counter == stoper) break;
    color[i] = (counter % k) + 1;
    counter++;
  }
  for (int ans : color) cout << ans << " ";
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tc;
  cin >> tc;
  for (int t = 1; t <= tc; t++) {
    solve();
  }
}

```