#include <bits/stdc++.h>
using namespace std;
long long int hcf(long long int a, long long int b) {
  if (b == 0) return a;
  return hcf(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / hcf(a, b);
}
bool isprime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
void solve() {
  long long int n, k;
  cin >> n >> k;
  vector<long long int> a;
  for (long long int i = 1; i <= n; i++) {
    long long int qq;
    cin >> qq;
    a.push_back(qq);
  }
  map<long long int, long long int> m;
  for (auto& x : a) m[x]++;
  vector<long long int> ans(n, 0);
  map<long long int, long long int> cnt;
  vector<long long int> temp;
  for (long long int i = 0; i <= n - 1; i++) {
    if (m[a[i]] >= k and cnt[a[i]] < k) {
      ans[i] = cnt[a[i]] + 1;
      cnt[a[i]]++;
    } else if (m[a[i]] < k)
      temp.push_back(a[i]);
  }
  if (temp.size() == 0) {
    for (auto& x : ans) cout << x << " ";
    cout << "\n";
    return;
  }
  sort(temp.begin(), temp.end());
  long long int c = temp.size() / k;
  map<long long int, vector<long long int> > final;
  for (long long int i = 0; i <= (c * k) - 1; i++) {
    long long int t = (i + 1) % k;
    if (t == 0) t = k;
    final[temp[i]].push_back(t);
  }
  for (long long int i = 0; i <= n - 1; i++) {
    if (ans[i] == 0) {
      if (final[a[i]].size() > 0) {
        ans[i] = final[a[i]].back();
        final[a[i]].pop_back();
      }
    }
  }
  for (auto& x : ans) cout << x << " ";
  cout << "\n";
}
void onlinejudge() {}
int main() {
  onlinejudge();
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int ttt;
  ttt = 1;
  cin >> ttt;
  cout << fixed << setprecision(16);
  while (ttt--) solve();
  return 0;
}

### Prompt

Develop a solution in cpp 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 hcf(long long int a, long long int b) {
  if (b == 0) return a;
  return hcf(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / hcf(a, b);
}
bool isprime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
void solve() {
  long long int n, k;
  cin >> n >> k;
  vector<long long int> a;
  for (long long int i = 1; i <= n; i++) {
    long long int qq;
    cin >> qq;
    a.push_back(qq);
  }
  map<long long int, long long int> m;
  for (auto& x : a) m[x]++;
  vector<long long int> ans(n, 0);
  map<long long int, long long int> cnt;
  vector<long long int> temp;
  for (long long int i = 0; i <= n - 1; i++) {
    if (m[a[i]] >= k and cnt[a[i]] < k) {
      ans[i] = cnt[a[i]] + 1;
      cnt[a[i]]++;
    } else if (m[a[i]] < k)
      temp.push_back(a[i]);
  }
  if (temp.size() == 0) {
    for (auto& x : ans) cout << x << " ";
    cout << "\n";
    return;
  }
  sort(temp.begin(), temp.end());
  long long int c = temp.size() / k;
  map<long long int, vector<long long int> > final;
  for (long long int i = 0; i <= (c * k) - 1; i++) {
    long long int t = (i + 1) % k;
    if (t == 0) t = k;
    final[temp[i]].push_back(t);
  }
  for (long long int i = 0; i <= n - 1; i++) {
    if (ans[i] == 0) {
      if (final[a[i]].size() > 0) {
        ans[i] = final[a[i]].back();
        final[a[i]].pop_back();
      }
    }
  }
  for (auto& x : ans) cout << x << " ";
  cout << "\n";
}
void onlinejudge() {}
int main() {
  onlinejudge();
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int ttt;
  ttt = 1;
  cin >> ttt;
  cout << fixed << setprecision(16);
  while (ttt--) solve();
  return 0;
}

```