#include <bits/stdc++.h>
using namespace std;
int max(int a, int b) { return a > b ? a : b; }
int min(int a, int b) { return a < b ? a : b; }
inline int read() {
  char ch = getchar();
  int x = 0, r = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') r = 0;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  return r ? x : -x;
}
int t, n, k, a[200100], b[200100], ans[200100], cnt[200100], l1, l2, top;
struct node {
  int k, no;
  bool operator<(const node b) const { return k < b.k; }
} q[200100];
int main() {
  t = read();
  while (t--) {
    n = read();
    k = read();
    for (int i = 1; i <= n; ++i) {
      a[i] = read();
      b[a[i]]++;
    }
    l1 = l2 = top = 0;
    for (int i = 1; i <= n; ++i) {
      if (b[a[i]] >= k)
        ans[i] = ++cnt[a[i]];
      else
        l1++, ans[i] = 0, q[++top].no = i, q[top].k = a[i];
    }
    l1 = (l1 / k) * k;
    sort(q + 1, q + top + 1);
    for (int i = 1; i <= top; ++i) {
      if (l2 >= l1) break;
      l2++;
      ans[q[i].no] = (l2 - 1) % k + 1;
    }
    for (int i = 1; i <= n; ++i) {
      if (ans[i] != 0 && ans[i] <= k)
        printf("%d ", ans[i]);
      else
        printf("0 ");
    }
    for (int i = 1; i <= n; ++i) a[i] = b[i] = ans[i] = cnt[i] = 0;
    puts("");
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
int max(int a, int b) { return a > b ? a : b; }
int min(int a, int b) { return a < b ? a : b; }
inline int read() {
  char ch = getchar();
  int x = 0, r = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') r = 0;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  return r ? x : -x;
}
int t, n, k, a[200100], b[200100], ans[200100], cnt[200100], l1, l2, top;
struct node {
  int k, no;
  bool operator<(const node b) const { return k < b.k; }
} q[200100];
int main() {
  t = read();
  while (t--) {
    n = read();
    k = read();
    for (int i = 1; i <= n; ++i) {
      a[i] = read();
      b[a[i]]++;
    }
    l1 = l2 = top = 0;
    for (int i = 1; i <= n; ++i) {
      if (b[a[i]] >= k)
        ans[i] = ++cnt[a[i]];
      else
        l1++, ans[i] = 0, q[++top].no = i, q[top].k = a[i];
    }
    l1 = (l1 / k) * k;
    sort(q + 1, q + top + 1);
    for (int i = 1; i <= top; ++i) {
      if (l2 >= l1) break;
      l2++;
      ans[q[i].no] = (l2 - 1) % k + 1;
    }
    for (int i = 1; i <= n; ++i) {
      if (ans[i] != 0 && ans[i] <= k)
        printf("%d ", ans[i]);
      else
        printf("0 ");
    }
    for (int i = 1; i <= n; ++i) a[i] = b[i] = ans[i] = cnt[i] = 0;
    puts("");
  }
  return 0;
}

```