#include <bits/stdc++.h>
using namespace std;
long long int fastpow(long long int a, long long int p) {
  long long int res = 1;
  while (p) {
    if (p % 2 == 0) {
      a = a * 1ll * a % 1000000007;
      p /= 2;
    } else {
      res = res * 1ll * a % 1000000007;
      p--;
    }
  }
  return res;
}
long long int fact(long long int n) {
  long long int res = 1;
  for (long long int i = 1; i <= n; i++) {
    res = res * 1ll * i % 1000000007;
  }
  return res;
}
long long int comb(long long int n, long long int k) {
  return fact(n) * 1ll * fastpow(fact(k), 1000000007 - 2) % 1000000007 * 1ll *
         fastpow(fact(n - k), 1000000007 - 2) % 1000000007;
}
long long int to_integer(string &aa) {
  long long int res = 0;
  for (auto x : aa) res = (res * 10) + (x - '0');
  return res;
}
bool prime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
void sieve(long long int n, vector<long long int> prime) {
  for (long long int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (long long int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int tt;
  cin >> tt;
  while (tt--) {
    long long int n, k;
    cin >> n >> k;
    map<long long int, vector<long long int> > pos;
    long long int cnt = 0;
    for (long long int i = 0; i < n; i++) {
      long long int x;
      cin >> x;
      if (pos[x].size() < k) {
        pos[x].push_back(i);
        cnt++;
      }
    }
    cnt -= cnt % k;
    vector<long long int> ans(n);
    for (long long int i = 1; i <= cnt / k; i++) {
      long long int have = 0;
      while (have < k) {
        auto it = pos.begin();
        if (it->second.empty()) {
          pos.erase(it);
          continue;
        }
        have++;
        ans[it->second.back()] = have;
        it->second.pop_back();
      }
    }
    for (auto i : ans) {
      cout << i << ' ';
    }
    cout << "\n";
  }
}

### 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 int fastpow(long long int a, long long int p) {
  long long int res = 1;
  while (p) {
    if (p % 2 == 0) {
      a = a * 1ll * a % 1000000007;
      p /= 2;
    } else {
      res = res * 1ll * a % 1000000007;
      p--;
    }
  }
  return res;
}
long long int fact(long long int n) {
  long long int res = 1;
  for (long long int i = 1; i <= n; i++) {
    res = res * 1ll * i % 1000000007;
  }
  return res;
}
long long int comb(long long int n, long long int k) {
  return fact(n) * 1ll * fastpow(fact(k), 1000000007 - 2) % 1000000007 * 1ll *
         fastpow(fact(n - k), 1000000007 - 2) % 1000000007;
}
long long int to_integer(string &aa) {
  long long int res = 0;
  for (auto x : aa) res = (res * 10) + (x - '0');
  return res;
}
bool prime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
void sieve(long long int n, vector<long long int> prime) {
  for (long long int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (long long int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int tt;
  cin >> tt;
  while (tt--) {
    long long int n, k;
    cin >> n >> k;
    map<long long int, vector<long long int> > pos;
    long long int cnt = 0;
    for (long long int i = 0; i < n; i++) {
      long long int x;
      cin >> x;
      if (pos[x].size() < k) {
        pos[x].push_back(i);
        cnt++;
      }
    }
    cnt -= cnt % k;
    vector<long long int> ans(n);
    for (long long int i = 1; i <= cnt / k; i++) {
      long long int have = 0;
      while (have < k) {
        auto it = pos.begin();
        if (it->second.empty()) {
          pos.erase(it);
          continue;
        }
        have++;
        ans[it->second.back()] = have;
        it->second.pop_back();
      }
    }
    for (auto i : ans) {
      cout << i << ' ';
    }
    cout << "\n";
  }
}

```