#include <bits/stdc++.h>
using namespace std;
long long t;
long long dd[200005];
struct hdz {
  long long i, j;
};
hdz A[200005];
bool cmp(hdz a, hdz b) { return a.j < b.j; }
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    long long a[n + 2];
    map<long long, long long> m;
    vector<long long> v[n + 2];
    set<long long> s;
    for (long long i = 1; i <= n; i++) {
      dd[i] = 0;
      A[i].i = 0;
      A[i].j = 0;
      cin >> a[i];
      m[a[i]]++;
      v[a[i]].push_back(i);
      s.insert(a[i]);
    }
    long long tmp = 0;
    for (long long i : s) {
      A[++tmp].i = i;
      A[tmp].j = m[i];
    }
    if (k == n) {
      for (long long i = 1; i <= n; i++) {
        cout << i << ' ';
      }
      cout << '\n';
      continue;
    }
    sort(A + 1, A + 1 + tmp, cmp);
    for (long long i = tmp; i >= 1; i--) {
      long long tmp = k;
      if (tmp > A[i].j) {
        long long d = 0;
        long long res = 0;
        bool kt = false;
        for (long long j = i; j >= 1; j--) {
          res += A[j].j;
          if (res >= tmp) {
            kt = true;
            break;
          }
        }
        long long j = i;
        if (kt)
          while (1) {
            if (tmp > A[j].j) {
              tmp -= A[j].j;
              A[j].j = 0;
              for (long long k = 0; k < v[A[j].i].size(); k++) {
                long long f = v[A[j].i][k];
                if (dd[f] != 0) continue;
                dd[f] = 1;
                a[f] = ++d;
              }
            } else {
              A[j].j = A[j].j - tmp;
              for (long long q = 0; q < v[A[j].i].size(); q++) {
                long long f = v[A[j].i][q];
                if (dd[f] != 0) continue;
                if (d == k) break;
                dd[f] = 1;
                a[f] = ++d;
              }
              cout << "\n\n";
              break;
            }
            j--;
          }
      } else {
        long long d = 0;
        for (long long j = 0; j < (long long)v[A[i].i].size(); j++) {
          if (d == k) break;
          long long f = v[A[i].i][j];
          if (dd[f] != 0) continue;
          dd[f] = 1;
          a[f] = ++d;
        }
      }
    }
    for (long long i = 1; i <= n; i++) {
      if (dd[i] != 0)
        cout << a[i] << ' ';
      else
        cout << 0 << ' ';
    }
    cout << '\n';
  }
}

### Prompt

Create a solution in CPP for the following problem:

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 t;
long long dd[200005];
struct hdz {
  long long i, j;
};
hdz A[200005];
bool cmp(hdz a, hdz b) { return a.j < b.j; }
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    long long a[n + 2];
    map<long long, long long> m;
    vector<long long> v[n + 2];
    set<long long> s;
    for (long long i = 1; i <= n; i++) {
      dd[i] = 0;
      A[i].i = 0;
      A[i].j = 0;
      cin >> a[i];
      m[a[i]]++;
      v[a[i]].push_back(i);
      s.insert(a[i]);
    }
    long long tmp = 0;
    for (long long i : s) {
      A[++tmp].i = i;
      A[tmp].j = m[i];
    }
    if (k == n) {
      for (long long i = 1; i <= n; i++) {
        cout << i << ' ';
      }
      cout << '\n';
      continue;
    }
    sort(A + 1, A + 1 + tmp, cmp);
    for (long long i = tmp; i >= 1; i--) {
      long long tmp = k;
      if (tmp > A[i].j) {
        long long d = 0;
        long long res = 0;
        bool kt = false;
        for (long long j = i; j >= 1; j--) {
          res += A[j].j;
          if (res >= tmp) {
            kt = true;
            break;
          }
        }
        long long j = i;
        if (kt)
          while (1) {
            if (tmp > A[j].j) {
              tmp -= A[j].j;
              A[j].j = 0;
              for (long long k = 0; k < v[A[j].i].size(); k++) {
                long long f = v[A[j].i][k];
                if (dd[f] != 0) continue;
                dd[f] = 1;
                a[f] = ++d;
              }
            } else {
              A[j].j = A[j].j - tmp;
              for (long long q = 0; q < v[A[j].i].size(); q++) {
                long long f = v[A[j].i][q];
                if (dd[f] != 0) continue;
                if (d == k) break;
                dd[f] = 1;
                a[f] = ++d;
              }
              cout << "\n\n";
              break;
            }
            j--;
          }
      } else {
        long long d = 0;
        for (long long j = 0; j < (long long)v[A[i].i].size(); j++) {
          if (d == k) break;
          long long f = v[A[i].i][j];
          if (dd[f] != 0) continue;
          dd[f] = 1;
          a[f] = ++d;
        }
      }
    }
    for (long long i = 1; i <= n; i++) {
      if (dd[i] != 0)
        cout << a[i] << ' ';
      else
        cout << 0 << ' ';
    }
    cout << '\n';
  }
}

```