#include <bits/stdc++.h>
using namespace std;
void fast(void) __attribute__((constructor));
void fast(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
template <typename T>
struct number_iterator : std::iterator<random_access_iterator_tag, T> {
  T v;
  number_iterator(T _v) : v(_v) {}
  operator T &() { return v; }
  T operator*() const { return v; }
};
template <typename T>
struct number_range {
  T b, e;
  number_range(T b, T e) : b(b), e(e) {}
  number_iterator<T> begin() { return b; }
  number_iterator<T> end() { return e; }
};
template <typename T>
number_range<T> range(T e) {
  return number_range<T>(0, e);
}
template <typename T>
number_range<T> range(T b, T e) {
  return number_range<T>(b, e);
}
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << endl;
  err(++it, args...);
}
inline void printV(const vector<long long> &v) {
  for_each(begin(v), end(v), [&](long long x) { cout << x << " "; });
}
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<vector<long long>> g(n + 1, vector<long long>());
  vector<long long> res(n, 0);
  for (long long i = 0; i < n; ++i) {
    long long x;
    cin >> x;
    g[x].emplace_back(i);
  }
  vector<long long> toColor;
  for (long long i = 1; i <= n; ++i) {
    if ((long long)g[i].size() >= k) {
      for (long long j = 0; j < k; ++j) {
        res[g[i][j]] = j + 1;
      }
    } else {
      for (long long j = 0; j < g[i].size(); ++j) {
        toColor.emplace_back(g[i][j]);
      }
    }
  }
  long long color = k;
  long long sz = k * (toColor.size() / k);
  for (long long i = 0; i < sz; ++i) {
    if (color == 0) color = k;
    res[toColor[i]] = color;
    --color;
  }
  for (auto &c : res) cout << c << " ";
}
signed main(void) {
  long long t;
  cin >> t;
  while (t--) {
    solve();
    cout << "\n";
  }
  return 0;
}

### Prompt

Your task is to create a cpp solution to 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 fast(void) __attribute__((constructor));
void fast(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
template <typename T>
struct number_iterator : std::iterator<random_access_iterator_tag, T> {
  T v;
  number_iterator(T _v) : v(_v) {}
  operator T &() { return v; }
  T operator*() const { return v; }
};
template <typename T>
struct number_range {
  T b, e;
  number_range(T b, T e) : b(b), e(e) {}
  number_iterator<T> begin() { return b; }
  number_iterator<T> end() { return e; }
};
template <typename T>
number_range<T> range(T e) {
  return number_range<T>(0, e);
}
template <typename T>
number_range<T> range(T b, T e) {
  return number_range<T>(b, e);
}
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << endl;
  err(++it, args...);
}
inline void printV(const vector<long long> &v) {
  for_each(begin(v), end(v), [&](long long x) { cout << x << " "; });
}
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<vector<long long>> g(n + 1, vector<long long>());
  vector<long long> res(n, 0);
  for (long long i = 0; i < n; ++i) {
    long long x;
    cin >> x;
    g[x].emplace_back(i);
  }
  vector<long long> toColor;
  for (long long i = 1; i <= n; ++i) {
    if ((long long)g[i].size() >= k) {
      for (long long j = 0; j < k; ++j) {
        res[g[i][j]] = j + 1;
      }
    } else {
      for (long long j = 0; j < g[i].size(); ++j) {
        toColor.emplace_back(g[i][j]);
      }
    }
  }
  long long color = k;
  long long sz = k * (toColor.size() / k);
  for (long long i = 0; i < sz; ++i) {
    if (color == 0) color = k;
    res[toColor[i]] = color;
    --color;
  }
  for (auto &c : res) cout << c << " ";
}
signed main(void) {
  long long t;
  cin >> t;
  while (t--) {
    solve();
    cout << "\n";
  }
  return 0;
}

```