#include <bits/stdc++.h>
using namespace std;
long long int fp(long long int n, long long int p) {
  if (p == 0) return 1;
  if (p == 1) return n;
  long long int res = fp(n, p / 2) % (long long)(1e9 + 7);
  res = ((res) * (res)) % (long long)(1e9 + 7);
  if (p % 2) res = (res * n) % (long long)(1e9 + 7);
  return res;
}
string tostring(long long int a) {
  string s = "";
  while (a > 0) {
    char c = a % 10 + '0';
    s.push_back(c);
    a /= 10;
  }
  reverse(s.begin(), s.end());
  return s;
}
int t, n, k, x, arr[200005];
int main() {
  ios::sync_with_stdio(0);
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  ;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    vector<vector<int>> v(n + 5);
    memset(arr, -1, sizeof(arr));
    map<int, int> mp;
    int cnt = 0, ind = 0, rem = 0, full = 0, c = 0;
    for (int i = 0; i < n; i++) {
      cin >> x;
      mp[x]++;
      v[x].push_back(i);
      if (mp[x] == k)
        full++;
      else if (mp[x] > k)
        rem++, arr[i] = -1;
    }
    c = n - (full * k) - rem;
    c = c - (c % k);
    for (auto i : mp) {
      if (i.second >= k) {
        for (int j = 0; j < k; j++) {
          arr[v[i.first][j]] = j + 1;
        }
      } else {
        for (int j = 0; j < v[i.first].size() && c > 0; j++) {
          arr[v[i.first][j]] = cnt + 1;
          cnt = (cnt + 1) % k;
          c--;
        }
      }
    }
    for (int i = 0; i < n; i++)
      if (arr[i] == -1)
        cout << 0 << " ";
      else
        cout << arr[i] << " ";
    cout << '\n';
  }
  return 0;
}

### 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;
long long int fp(long long int n, long long int p) {
  if (p == 0) return 1;
  if (p == 1) return n;
  long long int res = fp(n, p / 2) % (long long)(1e9 + 7);
  res = ((res) * (res)) % (long long)(1e9 + 7);
  if (p % 2) res = (res * n) % (long long)(1e9 + 7);
  return res;
}
string tostring(long long int a) {
  string s = "";
  while (a > 0) {
    char c = a % 10 + '0';
    s.push_back(c);
    a /= 10;
  }
  reverse(s.begin(), s.end());
  return s;
}
int t, n, k, x, arr[200005];
int main() {
  ios::sync_with_stdio(0);
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  ;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    vector<vector<int>> v(n + 5);
    memset(arr, -1, sizeof(arr));
    map<int, int> mp;
    int cnt = 0, ind = 0, rem = 0, full = 0, c = 0;
    for (int i = 0; i < n; i++) {
      cin >> x;
      mp[x]++;
      v[x].push_back(i);
      if (mp[x] == k)
        full++;
      else if (mp[x] > k)
        rem++, arr[i] = -1;
    }
    c = n - (full * k) - rem;
    c = c - (c % k);
    for (auto i : mp) {
      if (i.second >= k) {
        for (int j = 0; j < k; j++) {
          arr[v[i.first][j]] = j + 1;
        }
      } else {
        for (int j = 0; j < v[i.first].size() && c > 0; j++) {
          arr[v[i.first][j]] = cnt + 1;
          cnt = (cnt + 1) % k;
          c--;
        }
      }
    }
    for (int i = 0; i < n; i++)
      if (arr[i] == -1)
        cout << 0 << " ";
      else
        cout << arr[i] << " ";
    cout << '\n';
  }
  return 0;
}

```