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.
12
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
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
 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

2 1 2 3 0 1 3 2 3 1 
1 2 3 4 
1 
1 1 1 0 1 1 1 1 0 0 0 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 
2 1 2 3 0 1 3 2 3 1 
1 2 3 4 
1 
1 1 1 0 1 1 1 1 0 0 0 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 
 2 1 2 3 0 1 3 2 3 1 
1 2 3 4 
1 
1 1 1 0 1 1 1 1 0 0 0 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 

#include <bits/stdc++.h>
using namespace std;
namespace custom {
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, 1, 0, -1};
template <class T>
istream& operator>>(istream& in, vector<T>& v) {
  for (auto& el : v) in >> el;
  return in;
}
template <class T>
void print(T a, string end = "\n") {
  cout << a << end;
}
template <class T1, class T2>
void print(pair<T1, T2>& a, string end = "\n") {
  cout << "{" << a.first << "," << a.second << "}" << end;
}
template <class T>
void parr(T& arr, string end = "\n") {
  int i = 0;
  for (auto& el : arr) {
    print(el, " ");
  }
  print("", end);
}
template <class T>
void print(vector<T>& a, string end = "\n") {
  parr(a, end);
}
template <class T>
void print(set<T>& a, string end = "\n") {
  parr(a, end);
}
template <class T>
void print(deque<T>& a, string end = "\n") {
  parr(a, end);
}
template <class T1, class T2>
void print(map<T1, T2>& a, string end = "\n") {
  int i = 0;
  for (auto& el : a) {
    print(el.first, " : ");
    print(el.second);
  }
  print("", end);
}
template <class T>
void _print(T f, string end) {
  print(f, end);
}
template <class arg1, class... args>
void print(arg1 f, args... n) {
  _print(f, " ");
  print(n...);
}
template <class T>
void print(vector<vector<T>>& a) {
  for (auto& el : a) print(el, "\n");
}
template <class T>
int len(T& a) {
  return a.size();
}
template <class T>
long long sum(T& a) {
  long long s = 0;
  for (auto& el : a) s += el;
  return s;
}
template <class T1, class T2>
T1 min(T1 a, T2 b) {
  return (a < b) ? a : (T1)b;
}
template <class T1, class T2>
T1 max(T1 a, T2 b) {
  return (a > b) ? a : (T1)b;
}
template <class T>
T max(vector<T> a) {
  T m = a[0];
  for (auto& el : a) m = max(m, el);
  return m;
}
template <class T>
T min(vector<T> a) {
  T m = a[0];
  for (auto& el : a) m = min(m, el);
  return m;
}
template <class T1, class... T2>
T1 min(T1 a, T2... b) {
  return min(a, (T1)min(b...));
}
template <class T1, class... T2>
T1 max(T1 a, T2... b) {
  return max(a, (T1)max(b...));
}
}  // namespace custom
using namespace custom;
void solve() {
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  cin >> a;
  map<int, vector<int>> dic;
  for (int i = 0; i < n; i += 1) {
    dic[a[i]].push_back(i);
  }
  map<int, int> dic2;
  vector<int> temp;
  int j = 1;
  for (auto& el : dic) {
    if (el.second.size() >= k)
      dic2[el.first] = 1;
    else {
      for (auto& h : el.second) {
        temp.push_back(h);
      }
    }
  }
  vector<int> ans(n, 0);
  int o = temp.size() / k;
  if (o > 0) {
    int i = 0, c = 1;
    while (i < o * k) {
      for (int q = 0; q < o; q += 1) {
        ans[temp[i]] = c;
        i++;
        c += 1;
        if (c > k) c = 1;
      }
    }
  }
  for (int i = 0; i < n; i += 1) {
    if (ans[i] == 0) {
      if (dic2[a[i]] <= k) ans[i] = dic2[a[i]];
      dic2[a[i]] += 1;
    }
  }
  print(ans);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
