#include <bits/stdc++.h>
#pragma GCC optimize(3)
using namespace std;
inline int read() {
  int x = 0;
  int f = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    f |= (ch == '-') ? 1 : 0;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return f ? -x : x;
}
const int maxn = 2e5 + 10;
int n, t, k;
int aa[maxn], vis[maxn], vis1[maxn], id[maxn];
int cmp(int a, int b) { return aa[a] < aa[b]; }
int main() {
  t = read();
  while (t--) {
    n = read();
    k = read();
    for (int i = 1; i <= n; ++i) vis[i] = vis1[i] = 0, id[i] = i;
    for (int i = 1; i <= n; ++i) {
      aa[i] = read();
      ++vis[aa[i]];
    }
    int g = 0;
    for (int i = 1; i <= n; ++i)
      if (vis[i] > k)
        g += k;
      else
        g += vis[i];
    int u = g / k;
    sort(id + 1, id + n + 1, cmp);
    int tmp = 1, tmp1 = 1;
    int o = 1;
    for (int i = 1; i <= n; ++i)
      if (o <= u) {
        if (i == 1 || aa[id[i]] != aa[id[i - 1]])
          tmp = 1;
        else {
          ++tmp;
        }
        if (tmp <= k) {
          vis1[id[i]] = tmp1++;
          if (tmp1 > k) tmp1 = 1, ++o;
        }
      }
    for (int i = 1; i <= n; ++i) {
      if (i != 1) printf(" ");
      printf("%d", vis1[i]);
    }
    printf("\n");
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
#pragma GCC optimize(3)
using namespace std;
inline int read() {
  int x = 0;
  int f = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    f |= (ch == '-') ? 1 : 0;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return f ? -x : x;
}
const int maxn = 2e5 + 10;
int n, t, k;
int aa[maxn], vis[maxn], vis1[maxn], id[maxn];
int cmp(int a, int b) { return aa[a] < aa[b]; }
int main() {
  t = read();
  while (t--) {
    n = read();
    k = read();
    for (int i = 1; i <= n; ++i) vis[i] = vis1[i] = 0, id[i] = i;
    for (int i = 1; i <= n; ++i) {
      aa[i] = read();
      ++vis[aa[i]];
    }
    int g = 0;
    for (int i = 1; i <= n; ++i)
      if (vis[i] > k)
        g += k;
      else
        g += vis[i];
    int u = g / k;
    sort(id + 1, id + n + 1, cmp);
    int tmp = 1, tmp1 = 1;
    int o = 1;
    for (int i = 1; i <= n; ++i)
      if (o <= u) {
        if (i == 1 || aa[id[i]] != aa[id[i - 1]])
          tmp = 1;
        else {
          ++tmp;
        }
        if (tmp <= k) {
          vis1[id[i]] = tmp1++;
          if (tmp1 > k) tmp1 = 1, ++o;
        }
      }
    for (int i = 1; i <= n; ++i) {
      if (i != 1) printf(" ");
      printf("%d", vis1[i]);
    }
    printf("\n");
  }
}

```