#include <bits/stdc++.h>
using namespace std;
long long factorial(long long n) {
  long long res = 1, i;
  for (i = 2; i <= n; i++) {
    res *= i;
    res = res % 1000000007;
  }
  return res;
}
vector<long long> digits(long long n) {
  vector<long long> v;
  while (n) {
    v.push_back(n % 10);
    n /= 10;
  }
  return v;
}
vector<long long> SieveOfEratosthenes(long long n) {
  std::vector<long long> v(n + 1, 0);
  vector<long long> vb(n + 1, 1);
  long long i, j;
  std::vector<long long> ans;
  for (i = 2; i <= n; i++) {
    if (vb[i] == 1) {
      v[i] = i;
      ans.push_back(i);
      for (j = i * i; j <= n; j += i) {
        if (vb[i] == 1) {
          v[j] = i;
        }
        vb[j] = 0;
      }
    }
  }
  return ans;
}
map<long long, long long> get_prime_exponent(long long n) {
  std::vector<long long> v = SieveOfEratosthenes(n);
  long long i;
  long long cur = v[n];
  long long count = 0;
  map<long long, long long> m;
  while (n > 1) {
    cur = v[n];
    count = 0;
    while (n > 1 && n % cur == 0) {
      count++;
      n /= cur;
    }
    m[cur] = count;
  }
  return m;
}
void solve() {
  long long i, j, k, t, n;
  cin >> n >> j;
  std::vector<long long> v;
  for (i = 0; i <= n - 1; i++) {
    cin >> t;
    v.push_back(t);
  }
  std::vector<long long> vv = v;
  unordered_map<long long, long long> m;
  std::vector<std::vector<long long> > aa(n + 1);
  sort(v.begin(), v.end());
  long long color = 0;
  for (i = 0; i <= n - 1; i++) {
    if (m[v[i]] < j) {
      if (color == j) {
        color = 0;
      }
      aa[v[i]].push_back(color);
      color++;
    } else {
      aa[v[i]].push_back(-1);
    }
    m[v[i]]++;
  }
  std::vector<long long> colors(n, 0);
  if (color < j) {
    for (i = 0; i <= color - 1; i++) colors[i] = 1;
  }
  for (i = 0; i <= n - 1; i++) {
    t = aa[vv[i]].back();
    aa[vv[i]].pop_back();
    if (t == -1) {
      cout << 0;
    } else if (colors[t] == 0) {
      cout << t + 1;
    } else if (colors[t]) {
      cout << 0;
      colors[t]--;
    } else
      cout << 0;
    cout << " ";
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long n;
  cin >> n;
  while (n--) {
    solve();
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 factorial(long long n) {
  long long res = 1, i;
  for (i = 2; i <= n; i++) {
    res *= i;
    res = res % 1000000007;
  }
  return res;
}
vector<long long> digits(long long n) {
  vector<long long> v;
  while (n) {
    v.push_back(n % 10);
    n /= 10;
  }
  return v;
}
vector<long long> SieveOfEratosthenes(long long n) {
  std::vector<long long> v(n + 1, 0);
  vector<long long> vb(n + 1, 1);
  long long i, j;
  std::vector<long long> ans;
  for (i = 2; i <= n; i++) {
    if (vb[i] == 1) {
      v[i] = i;
      ans.push_back(i);
      for (j = i * i; j <= n; j += i) {
        if (vb[i] == 1) {
          v[j] = i;
        }
        vb[j] = 0;
      }
    }
  }
  return ans;
}
map<long long, long long> get_prime_exponent(long long n) {
  std::vector<long long> v = SieveOfEratosthenes(n);
  long long i;
  long long cur = v[n];
  long long count = 0;
  map<long long, long long> m;
  while (n > 1) {
    cur = v[n];
    count = 0;
    while (n > 1 && n % cur == 0) {
      count++;
      n /= cur;
    }
    m[cur] = count;
  }
  return m;
}
void solve() {
  long long i, j, k, t, n;
  cin >> n >> j;
  std::vector<long long> v;
  for (i = 0; i <= n - 1; i++) {
    cin >> t;
    v.push_back(t);
  }
  std::vector<long long> vv = v;
  unordered_map<long long, long long> m;
  std::vector<std::vector<long long> > aa(n + 1);
  sort(v.begin(), v.end());
  long long color = 0;
  for (i = 0; i <= n - 1; i++) {
    if (m[v[i]] < j) {
      if (color == j) {
        color = 0;
      }
      aa[v[i]].push_back(color);
      color++;
    } else {
      aa[v[i]].push_back(-1);
    }
    m[v[i]]++;
  }
  std::vector<long long> colors(n, 0);
  if (color < j) {
    for (i = 0; i <= color - 1; i++) colors[i] = 1;
  }
  for (i = 0; i <= n - 1; i++) {
    t = aa[vv[i]].back();
    aa[vv[i]].pop_back();
    if (t == -1) {
      cout << 0;
    } else if (colors[t] == 0) {
      cout << t + 1;
    } else if (colors[t]) {
      cout << 0;
      colors[t]--;
    } else
      cout << 0;
    cout << " ";
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long n;
  cin >> n;
  while (n--) {
    solve();
  }
  return 0;
}

```