#include <bits/stdc++.h>
using namespace std;
long long ceildiv(long long one, long long two) {
  if (one % two == 0) {
    return one / two;
  } else {
    return one / two + 1;
  }
}
long long power(long long n, long long pow, long long m) {
  if (pow == 0) return 1;
  if (pow % 2 == 0) {
    long long x = power(n, pow / 2, m);
    return (x * x) % m;
  } else
    return (power(n, pow - 1, m) * n) % m;
}
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
long long factorial(long long n, long long mod) {
  if (n > 1)
    return (n * factorial(n - 1, mod)) % mod;
  else
    return 1;
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
vector<long long> read(long long n) {
  vector<long long> a;
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    a.push_back(x);
  }
  return a;
}
vector<vector<pair<long long, long long>>> adj;
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long> a = read(n);
  vector<long long> rep = a;
  sort(a.begin(), a.end());
  vector<vector<long long>> colors(k);
  queue<long long> q;
  for (long long i = 0; i < k; i++) {
    q.push(i);
  }
  for (long long i = 0; i < n; i++) {
    long long ind = q.front();
    long long x = upper_bound(colors[ind].begin(), colors[ind].end(), a[i]) -
                  lower_bound(colors[ind].begin(), colors[ind].end(), a[i]);
    if (x == 0) {
      colors[ind].push_back(a[i]);
      q.pop();
      q.push(ind);
    } else {
      continue;
    }
  }
  long long len = 1000000000;
  for (long long i = 0; i < k; i++) {
    len = min(len, (long long)colors[i].size());
  }
  for (long long i = 0; i < k; i++) {
    while (colors[i].size() > len) {
      colors[i].pop_back();
    }
  }
  map<long long, vector<long long>> mp;
  for (long long i = 0; i < k; i++) {
    for (auto x : colors[i]) {
      mp[x].push_back(i + 1);
    }
  }
  for (long long i = 0; i < n; i++) {
    if (mp[rep[i]].size() == 0) {
      cout << 0 << " ";
    } else {
      cout << mp[rep[i]].back() << " ";
      mp[rep[i]].pop_back();
    }
  }
  cout << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  ;
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Please create a solution in CPP 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;
long long ceildiv(long long one, long long two) {
  if (one % two == 0) {
    return one / two;
  } else {
    return one / two + 1;
  }
}
long long power(long long n, long long pow, long long m) {
  if (pow == 0) return 1;
  if (pow % 2 == 0) {
    long long x = power(n, pow / 2, m);
    return (x * x) % m;
  } else
    return (power(n, pow - 1, m) * n) % m;
}
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
long long factorial(long long n, long long mod) {
  if (n > 1)
    return (n * factorial(n - 1, mod)) % mod;
  else
    return 1;
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
vector<long long> read(long long n) {
  vector<long long> a;
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    a.push_back(x);
  }
  return a;
}
vector<vector<pair<long long, long long>>> adj;
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long> a = read(n);
  vector<long long> rep = a;
  sort(a.begin(), a.end());
  vector<vector<long long>> colors(k);
  queue<long long> q;
  for (long long i = 0; i < k; i++) {
    q.push(i);
  }
  for (long long i = 0; i < n; i++) {
    long long ind = q.front();
    long long x = upper_bound(colors[ind].begin(), colors[ind].end(), a[i]) -
                  lower_bound(colors[ind].begin(), colors[ind].end(), a[i]);
    if (x == 0) {
      colors[ind].push_back(a[i]);
      q.pop();
      q.push(ind);
    } else {
      continue;
    }
  }
  long long len = 1000000000;
  for (long long i = 0; i < k; i++) {
    len = min(len, (long long)colors[i].size());
  }
  for (long long i = 0; i < k; i++) {
    while (colors[i].size() > len) {
      colors[i].pop_back();
    }
  }
  map<long long, vector<long long>> mp;
  for (long long i = 0; i < k; i++) {
    for (auto x : colors[i]) {
      mp[x].push_back(i + 1);
    }
  }
  for (long long i = 0; i < n; i++) {
    if (mp[rep[i]].size() == 0) {
      cout << 0 << " ";
    } else {
      cout << mp[rep[i]].back() << " ";
      mp[rep[i]].pop_back();
    }
  }
  cout << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  ;
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

```