#include <bits/stdc++.h>
using namespace std;
vector<int> v;
long long flag[10000000 + 1];
void sieve() {
  for (long long i = 1; i <= 10000000; i++) flag[i] = 1;
  flag[1] = 0;
  for (long long i = 2; i * i <= 10000000; i++) {
    if (flag[i] == 1) {
      for (long long j = i * i; j <= 10000000; j = j + i) flag[j] = 0;
    }
  }
}
long long bigMod(long long a, long long b, long long M) {
  if (b == 0) return 1 % M;
  long long x = bigMod(a, b / 2, M);
  x = ((x % M) * (x % M)) % M;
  if (b % 2 == 1) {
    long long y = ((x % M) * (a % M)) % M;
    return y;
  }
  return x;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    vector<long long> v1, v2, v3, v4, v5;
    vector<pair<long long, long long>> vpp, vp2;
    set<long long> st;
    set<long long>::iterator it;
    map<long long, long long> mp;
    long long k;
    cin >> n >> k;
    vector<long long> vp[n + 1];
    long long arr[n];
    memset(arr, 0, sizeof(arr));
    long long sum = 0, x;
    for (long long i = 0; i < n; i++) {
      cin >> x;
      mp[x]++;
      vp[x].push_back(i);
    }
    long long number = 1, index;
    for (auto it : mp) {
      long long number = 1;
      long long a = it.first;
      long long b = it.second;
      if (it.second >= k) {
        for (long long i = 0; i < vp[a].size(); i++) {
          index = vp[a][i];
          arr[index] = number;
          number++;
          if (number > k) break;
        }
      } else {
        for (long long i = 0; i < vp[a].size(); i++)
          vp2.push_back(make_pair(a, vp[a][i]));
      }
    }
    long long baki = vp2.size() - (vp2.size() % k);
    x = 1;
    for (long long i = 0; i < baki; i++) {
      index = vp2[i].second;
      arr[index] = x;
      x++;
      if (x > k) x = 1;
    }
    for (long long i = 0; i < n; i++) cout << arr[i] << " ";
    cout << endl;
  }
  return 0;
}

### 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;
vector<int> v;
long long flag[10000000 + 1];
void sieve() {
  for (long long i = 1; i <= 10000000; i++) flag[i] = 1;
  flag[1] = 0;
  for (long long i = 2; i * i <= 10000000; i++) {
    if (flag[i] == 1) {
      for (long long j = i * i; j <= 10000000; j = j + i) flag[j] = 0;
    }
  }
}
long long bigMod(long long a, long long b, long long M) {
  if (b == 0) return 1 % M;
  long long x = bigMod(a, b / 2, M);
  x = ((x % M) * (x % M)) % M;
  if (b % 2 == 1) {
    long long y = ((x % M) * (a % M)) % M;
    return y;
  }
  return x;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    vector<long long> v1, v2, v3, v4, v5;
    vector<pair<long long, long long>> vpp, vp2;
    set<long long> st;
    set<long long>::iterator it;
    map<long long, long long> mp;
    long long k;
    cin >> n >> k;
    vector<long long> vp[n + 1];
    long long arr[n];
    memset(arr, 0, sizeof(arr));
    long long sum = 0, x;
    for (long long i = 0; i < n; i++) {
      cin >> x;
      mp[x]++;
      vp[x].push_back(i);
    }
    long long number = 1, index;
    for (auto it : mp) {
      long long number = 1;
      long long a = it.first;
      long long b = it.second;
      if (it.second >= k) {
        for (long long i = 0; i < vp[a].size(); i++) {
          index = vp[a][i];
          arr[index] = number;
          number++;
          if (number > k) break;
        }
      } else {
        for (long long i = 0; i < vp[a].size(); i++)
          vp2.push_back(make_pair(a, vp[a][i]));
      }
    }
    long long baki = vp2.size() - (vp2.size() % k);
    x = 1;
    for (long long i = 0; i < baki; i++) {
      index = vp2[i].second;
      arr[index] = x;
      x++;
      if (x > k) x = 1;
    }
    for (long long i = 0; i < n; i++) cout << arr[i] << " ";
    cout << endl;
  }
  return 0;
}

```