#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
const int mod = 1e9 + 7;
int t, n, k;
int a[maxn], num[maxn];
int NUM[maxn], vis[maxn];
struct node {
  int val;
  int pos;
  int color;
} it[maxn];
struct nnode {
  int l, r;
} pos[maxn];
bool cmp1(node a, node b) {
  if (num[a.val] != num[b.val])
    return num[a.val] > num[b.val];
  else if (a.val != b.val)
    return a.val < b.val;
  else
    return a.pos < b.pos;
}
bool cmp2(node a, node b) { return a.pos < b.pos; }
int main() {
  cin >> t;
  while (t--) {
    cin >> n >> k;
    int ans = 0;
    memset(NUM, 0, sizeof(NUM));
    memset(vis, 0, sizeof(vis));
    memset(pos, 0, sizeof(pos));
    int nnum = 0;
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      it[i].val = a[i];
      it[i].pos = i;
      it[i].color = 0;
      num[i] = 0;
    }
    for (int i = 1; i <= n; i++) num[a[i]]++;
    for (int i = 1; i <= n; i++) {
      if (num[i]) {
        NUM[nnum++] = i;
        ans += min(num[i], k);
      }
    }
    int cnt = ans / k;
    sort(it + 1, it + 1 + n, cmp1);
    int all = cnt * k;
    int last = 0;
    for (int i = 1; i <= n; i++) {
      int nn = num[it[i].val], nnn = it[i].val;
      if (nn >= k) {
        int col = 1, j;
        for (j = i; j < i + k; j++) it[j].color = col++;
        while (it[j].val == nnn) j++;
        i = j - 1;
        all -= k;
      } else {
        last = i;
        break;
      }
    }
    for (int i = last; i <= n; i++) {
      if (all == 0) break;
      if ((i - last + 1) % k == 0)
        it[i].color = k;
      else
        it[i].color = (i - last + 1) % k;
      all--;
    }
    sort(it + 1, it + 1 + n, cmp2);
    for (int i = 1; i <= n; i++) {
      printf("%d ", it[i].color);
    }
    printf("\n");
  }
  return 0;
}

### Prompt

Please formulate a cpp solution 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;
const int maxn = 2e5 + 5;
const int mod = 1e9 + 7;
int t, n, k;
int a[maxn], num[maxn];
int NUM[maxn], vis[maxn];
struct node {
  int val;
  int pos;
  int color;
} it[maxn];
struct nnode {
  int l, r;
} pos[maxn];
bool cmp1(node a, node b) {
  if (num[a.val] != num[b.val])
    return num[a.val] > num[b.val];
  else if (a.val != b.val)
    return a.val < b.val;
  else
    return a.pos < b.pos;
}
bool cmp2(node a, node b) { return a.pos < b.pos; }
int main() {
  cin >> t;
  while (t--) {
    cin >> n >> k;
    int ans = 0;
    memset(NUM, 0, sizeof(NUM));
    memset(vis, 0, sizeof(vis));
    memset(pos, 0, sizeof(pos));
    int nnum = 0;
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      it[i].val = a[i];
      it[i].pos = i;
      it[i].color = 0;
      num[i] = 0;
    }
    for (int i = 1; i <= n; i++) num[a[i]]++;
    for (int i = 1; i <= n; i++) {
      if (num[i]) {
        NUM[nnum++] = i;
        ans += min(num[i], k);
      }
    }
    int cnt = ans / k;
    sort(it + 1, it + 1 + n, cmp1);
    int all = cnt * k;
    int last = 0;
    for (int i = 1; i <= n; i++) {
      int nn = num[it[i].val], nnn = it[i].val;
      if (nn >= k) {
        int col = 1, j;
        for (j = i; j < i + k; j++) it[j].color = col++;
        while (it[j].val == nnn) j++;
        i = j - 1;
        all -= k;
      } else {
        last = i;
        break;
      }
    }
    for (int i = last; i <= n; i++) {
      if (all == 0) break;
      if ((i - last + 1) % k == 0)
        it[i].color = k;
      else
        it[i].color = (i - last + 1) % k;
      all--;
    }
    sort(it + 1, it + 1 + n, cmp2);
    for (int i = 1; i <= n; i++) {
      printf("%d ", it[i].color);
    }
    printf("\n");
  }
  return 0;
}

```