#include <bits/stdc++.h>
using namespace std;
long long power(long long x, long long y);
bool isPrime(long long n);
long long modInv(long long a, long long b);
long long gcdExtended(long long a, long long b, long long *x, long long *y);
long long mpower(long long a, long long b, long long p);
bool sorted(vector<long long> v, long long n, long long strict);
const long long inf = (long long)1e17;
const long long N = (long long)3e5 + 5;
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long> v(n);
  map<long long, vector<long long> > same;
  long long tot = 0;
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
    if (same[v[i]].size() < k) same[v[i]].push_back(i);
  }
  vector<long long> colored_(n, 0);
  for (auto color : same) {
    tot += color.second.size();
  }
  tot = k * (tot / k);
  long long current_color = 1;
  for (auto color : same) {
    for (auto positions : color.second) {
      colored_[positions] = current_color;
      current_color += 1;
      if (current_color > k) current_color = 1;
      tot -= 1;
      if (tot == 0) break;
    }
    if (tot == 0) break;
  }
  for (auto i : colored_) cout << i << ' ';
  ;
  cout << endl;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  clock_t start, end;
  start = clock();
  long long x;
  cin >> x;
  while (x--) {
    solve();
  };
  end = clock();
  double time_taken = double(end - start) / (double)CLOCKS_PER_SEC;
  return 0;
}
long long modInv(long long a, long long m) {
  long long x, y;
  long long g = gcdExtended(a, m, &x, &y);
  long long res = (x % m + m) % m;
  return res;
}
long long gcdExtended(long long a, long long b, long long *x, long long *y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  long long x1, y1;
  long long gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
long long mpower(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = res * x;
    y = y >> 1;
    x = x * x;
  }
  return res;
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  long long p = sqrt(n);
  for (long long i = 5; i <= p; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
bool sorted(vector<long long> v, long long n, long long strict) {
  if (n <= 1) return 1;
  for (long long i = 1; i < n; i++) {
    if (strict) {
      if (v[i] <= v[i - 1]) return 0;
    } else {
      if (v[i] < v[i - 1]) return 0;
    }
  }
  return 1;
}

### 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 power(long long x, long long y);
bool isPrime(long long n);
long long modInv(long long a, long long b);
long long gcdExtended(long long a, long long b, long long *x, long long *y);
long long mpower(long long a, long long b, long long p);
bool sorted(vector<long long> v, long long n, long long strict);
const long long inf = (long long)1e17;
const long long N = (long long)3e5 + 5;
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long> v(n);
  map<long long, vector<long long> > same;
  long long tot = 0;
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
    if (same[v[i]].size() < k) same[v[i]].push_back(i);
  }
  vector<long long> colored_(n, 0);
  for (auto color : same) {
    tot += color.second.size();
  }
  tot = k * (tot / k);
  long long current_color = 1;
  for (auto color : same) {
    for (auto positions : color.second) {
      colored_[positions] = current_color;
      current_color += 1;
      if (current_color > k) current_color = 1;
      tot -= 1;
      if (tot == 0) break;
    }
    if (tot == 0) break;
  }
  for (auto i : colored_) cout << i << ' ';
  ;
  cout << endl;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  clock_t start, end;
  start = clock();
  long long x;
  cin >> x;
  while (x--) {
    solve();
  };
  end = clock();
  double time_taken = double(end - start) / (double)CLOCKS_PER_SEC;
  return 0;
}
long long modInv(long long a, long long m) {
  long long x, y;
  long long g = gcdExtended(a, m, &x, &y);
  long long res = (x % m + m) % m;
  return res;
}
long long gcdExtended(long long a, long long b, long long *x, long long *y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  long long x1, y1;
  long long gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
long long mpower(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = res * x;
    y = y >> 1;
    x = x * x;
  }
  return res;
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  long long p = sqrt(n);
  for (long long i = 5; i <= p; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
bool sorted(vector<long long> v, long long n, long long strict) {
  if (n <= 1) return 1;
  for (long long i = 1; i < n; i++) {
    if (strict) {
      if (v[i] <= v[i - 1]) return 0;
    } else {
      if (v[i] < v[i - 1]) return 0;
    }
  }
  return 1;
}

```