#include <bits/stdc++.h>
using namespace std;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << '\n';
  err(++it, args...);
}
using ll = long long;
using ld = long double;
using vi = vector<int>;
using vii = vector<vi>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
const int MOD = 1000000007;
const ll INF = 1000000007;
const int V = 200005;
int cnt[V];
int a[V];
int color[V];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    memset(cnt, 0, sizeof(cnt));
    for (int i = 0; i < n; ++i) cin >> a[i], --a[i], ++cnt[a[i]];
    int C = 0;
    for (int i = 0; i < n; ++i)
      if (0 < cnt[i] && cnt[i] < k) C += cnt[i];
    C = C / k * k;
    memset(color, 0, sizeof(color));
    int cc = 0;
    for (int i = 0; i < n; ++i)
      if (0 < cnt[i] && cnt[i] < k) color[i] = cc, cc += cnt[i];
    int c = -1;
    for (int i = 0; i < n; ++i) {
      if (cnt[a[i]] >= k) {
        if (color[a[i]] == k) {
          cout << 0 << ' ';
          continue;
        }
        cout << ++color[a[i]] << ' ';
      } else {
        if (color[a[i]] < C) {
          cout << ((color[a[i]]++) % k) + 1 << ' ';
          continue;
        }
        cout << 0 << ' ';
      }
    }
    cout << '\n';
  }
  return 0;
}

### 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;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << '\n';
  err(++it, args...);
}
using ll = long long;
using ld = long double;
using vi = vector<int>;
using vii = vector<vi>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
const int MOD = 1000000007;
const ll INF = 1000000007;
const int V = 200005;
int cnt[V];
int a[V];
int color[V];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    memset(cnt, 0, sizeof(cnt));
    for (int i = 0; i < n; ++i) cin >> a[i], --a[i], ++cnt[a[i]];
    int C = 0;
    for (int i = 0; i < n; ++i)
      if (0 < cnt[i] && cnt[i] < k) C += cnt[i];
    C = C / k * k;
    memset(color, 0, sizeof(color));
    int cc = 0;
    for (int i = 0; i < n; ++i)
      if (0 < cnt[i] && cnt[i] < k) color[i] = cc, cc += cnt[i];
    int c = -1;
    for (int i = 0; i < n; ++i) {
      if (cnt[a[i]] >= k) {
        if (color[a[i]] == k) {
          cout << 0 << ' ';
          continue;
        }
        cout << ++color[a[i]] << ' ';
      } else {
        if (color[a[i]] < C) {
          cout << ((color[a[i]]++) % k) + 1 << ' ';
          continue;
        }
        cout << 0 << ' ';
      }
    }
    cout << '\n';
  }
  return 0;
}

```