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;
const long long mod = 1e9 + 17;
const long long inf = 1e18 + 1;
long long modi(long long numberToMod) {
  numberToMod += mod;
  numberToMod %= mod;
  return numberToMod;
}
long long powerWithMod(long long numberToPower, long long myPower) {
  if (!myPower) return long long(1);
  long long recursive = powerWithMod(numberToPower, myPower / long long(2));
  if (myPower % 2) return modi(modi(recursive * recursive) * numberToPower);
  return modi(recursive * recursive);
}
long long power(long long numberToPower, long long myPower) {
  if (!myPower) return long long(1);
  long long recursive = powerWithMod(numberToPower, myPower / long long(2));
  if (myPower % 2) return recursive * recursive * numberToPower;
  return recursive * recursive;
}
long long gcd(long long numberOne, long long numberTwo) {
  if (numberTwo == 0) return numberOne;
  return gcd(numberTwo, numberOne % numberTwo);
}
vector<long long> compute_pi(string pattern) {
  vector<long long> result(pattern.size());
  result[0] = 0;
  for (unsigned int i = 1; i < pattern.size(); i++) {
    result[i] = 0;
    int KMP_count = result[i - 1];
    while (KMP_count > 0 && pattern[KMP_count] != pattern[i])
      KMP_count = result[KMP_count - 1];
    if (pattern[KMP_count] == pattern[i]) result[i] = KMP_count + 1;
  }
  return result;
}
const int maxn = 2e5 + 10;
long long t;
long long m, x, k, a[maxn], n, c[maxn];
vector<int> v[maxn], save;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    cin >> n >> k;
    for (int i = 1; i <= n; i++) v[i].clear();
    for (int i = 0; i < n; i++) c[i] = 0;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      v[a[i]].push_back(i);
    }
    m = 0;
    x = 0;
    save.clear();
    for (int i = 1; i <= n; i++) {
      if (v[i].size() >= k) {
        for (int i2 = 0; i2 < k; i2++) c[v[i][i2]] = i2 + 1;
      } else {
        for (int i2 = 0; i2 < v[i].size(); i2++) save.push_back(v[i][i2]);
      }
    }
    x = int(save.size()) % k;
    for (int i = 0; i < save.size() - x; i++) c[save[i]] = (i % k) + 1;
    for (int i = 0; i < n; i++) cout << c[i] << " ";
    cout << endl;
  }
  return 0;
}
