#include <bits/stdc++.h>
using namespace std;
template <typename T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <typename T>
T lcm(T a, T b) {
  return a * b / gcd(a, b);
}
const int mod = 998244353;
const int MOD = 1e9 + 7;
inline long long read() {
  long long ans = 0;
  char ch = getchar(), las = ' ';
  while (!isdigit(ch)) las = ch, ch = getchar();
  while (isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if (las == '-') ans = -ans;
  return ans;
}
inline void write(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
const int N = 4e5;
int b[N];
void run_case() {
  int n = read(), k = read();
  map<int, int> m;
  map<int, deque<int> > g;
  for (int i = (1); i <= (n); i++) {
    b[i] = read();
    m[b[i]]++;
  }
  int f = 0;
  map<int, int>::iterator it;
  for (it = m.begin(); it != m.end(); it++) {
    if (it->second == k) {
      for (int j = 1; j <= it->second; j++) {
        g[it->first].push_back((j % k == 0) ? k : j % k);
      }
    } else if (it->second > k) {
      for (int j = 1; j <= it->second; j++) {
        if (j <= k)
          g[it->first].push_back((j % k == 0) ? k : j % k);
        else
          g[it->first].push_back(0);
      }
    } else {
      for (int j = 1; j <= it->second; j++) {
        f++;
        g[it->first].push_back((f % k == 0) ? k : f % k);
      }
    }
  }
  int ff, tt;
  map<int, int>::reverse_iterator iter;
  if (f % k) {
    for (iter = m.rbegin(); iter != m.rend(); iter++) {
      if (iter->second < k) {
        ff = f;
        for (int j = 1; j <= iter->second; j++) {
          f--;
          g[iter->first].pop_back();
          if (f % k == 0) {
            tt = iter->first;
            goto end;
          }
        }
        for (int j = 1; j <= iter->second; j++) {
          g[iter->first].push_back(0);
        }
      }
    }
  end:
    for (int i = 1; i <= ff - f; i++) g[tt].push_back(0);
  }
  for (int i = 1; i <= n; i++) {
    if (i != 1) printf(" ");
    printf("%d", g[b[i]].front());
    g[b[i]].pop_front();
  }
  puts("");
}
int main() {
  int _ = read();
  while (_--) {
    run_case();
  }
}

### 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;
template <typename T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <typename T>
T lcm(T a, T b) {
  return a * b / gcd(a, b);
}
const int mod = 998244353;
const int MOD = 1e9 + 7;
inline long long read() {
  long long ans = 0;
  char ch = getchar(), las = ' ';
  while (!isdigit(ch)) las = ch, ch = getchar();
  while (isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if (las == '-') ans = -ans;
  return ans;
}
inline void write(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
const int N = 4e5;
int b[N];
void run_case() {
  int n = read(), k = read();
  map<int, int> m;
  map<int, deque<int> > g;
  for (int i = (1); i <= (n); i++) {
    b[i] = read();
    m[b[i]]++;
  }
  int f = 0;
  map<int, int>::iterator it;
  for (it = m.begin(); it != m.end(); it++) {
    if (it->second == k) {
      for (int j = 1; j <= it->second; j++) {
        g[it->first].push_back((j % k == 0) ? k : j % k);
      }
    } else if (it->second > k) {
      for (int j = 1; j <= it->second; j++) {
        if (j <= k)
          g[it->first].push_back((j % k == 0) ? k : j % k);
        else
          g[it->first].push_back(0);
      }
    } else {
      for (int j = 1; j <= it->second; j++) {
        f++;
        g[it->first].push_back((f % k == 0) ? k : f % k);
      }
    }
  }
  int ff, tt;
  map<int, int>::reverse_iterator iter;
  if (f % k) {
    for (iter = m.rbegin(); iter != m.rend(); iter++) {
      if (iter->second < k) {
        ff = f;
        for (int j = 1; j <= iter->second; j++) {
          f--;
          g[iter->first].pop_back();
          if (f % k == 0) {
            tt = iter->first;
            goto end;
          }
        }
        for (int j = 1; j <= iter->second; j++) {
          g[iter->first].push_back(0);
        }
      }
    }
  end:
    for (int i = 1; i <= ff - f; i++) g[tt].push_back(0);
  }
  for (int i = 1; i <= n; i++) {
    if (i != 1) printf(" ");
    printf("%d", g[b[i]].front());
    g[b[i]].pop_front();
  }
  puts("");
}
int main() {
  int _ = read();
  while (_--) {
    run_case();
  }
}

```