#include <bits/stdc++.h>
using namespace std;
int const N = 1e9 + 7;
vector<long long> seive(1000000, 1);
vector<long long> primes;
void doSeive() {
  for (long long i = 2; i < 1000000; i++) {
    if (seive[i]) {
      primes.push_back(i);
      for (long long j = i * i; j < 1000000; j += i) {
        seive[j] = 0;
      }
    }
  }
}
long long nCr(long long n, long long r) {
  if (n < r) return 0;
  if (r > n - r) {
    r = n - r;
  }
  long long res = 1;
  for (long long i = 0; i < r; i++) {
    res *= n - i;
    res /= i + 1;
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return (b % a, a);
}
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long> adj[n + 1];
  vector<long long> col(n, 0);
  for (long long i = 0; i < n; i++) {
    long long no;
    cin >> no;
    adj[no].push_back(i);
  }
  vector<long long> vis(n + 1, 0);
  long long cnt = n;
  for (long long i = 1; i <= n; i++) {
    if (adj[i].size() >= k) {
      vis[i] = 1;
      cnt -= adj[i].size();
      for (int j = 0; j < k; j++) {
        col[adj[i][j]] = j + 1;
      }
    }
  }
  cnt -= (cnt % k);
  long long p = 0;
  for (long long i = 1; i <= n; i++) {
    if (vis[i] == 0 && adj[i].size() > 0) {
      for (int j = 0; j < adj[i].size() && cnt > 0; j++) {
        p++;
        if (p == k + 1) p = 1;
        col[adj[i][j]] = p;
        cnt--;
      }
    }
  }
  for (long long i = 0; i < n; i++) {
    cout << col[i] << " ";
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
int const N = 1e9 + 7;
vector<long long> seive(1000000, 1);
vector<long long> primes;
void doSeive() {
  for (long long i = 2; i < 1000000; i++) {
    if (seive[i]) {
      primes.push_back(i);
      for (long long j = i * i; j < 1000000; j += i) {
        seive[j] = 0;
      }
    }
  }
}
long long nCr(long long n, long long r) {
  if (n < r) return 0;
  if (r > n - r) {
    r = n - r;
  }
  long long res = 1;
  for (long long i = 0; i < r; i++) {
    res *= n - i;
    res /= i + 1;
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return (b % a, a);
}
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long> adj[n + 1];
  vector<long long> col(n, 0);
  for (long long i = 0; i < n; i++) {
    long long no;
    cin >> no;
    adj[no].push_back(i);
  }
  vector<long long> vis(n + 1, 0);
  long long cnt = n;
  for (long long i = 1; i <= n; i++) {
    if (adj[i].size() >= k) {
      vis[i] = 1;
      cnt -= adj[i].size();
      for (int j = 0; j < k; j++) {
        col[adj[i][j]] = j + 1;
      }
    }
  }
  cnt -= (cnt % k);
  long long p = 0;
  for (long long i = 1; i <= n; i++) {
    if (vis[i] == 0 && adj[i].size() > 0) {
      for (int j = 0; j < adj[i].size() && cnt > 0; j++) {
        p++;
        if (p == k + 1) p = 1;
        col[adj[i][j]] = p;
        cnt--;
      }
    }
  }
  for (long long i = 0; i < n; i++) {
    cout << col[i] << " ";
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

```