B2. Wonderful Coloring - 2

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 a1,a2,…,ana1,a2,…,an. They want to paint it using pieces of chalk of kk 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 kk 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 kk 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]a=[3,1,1,1,1,10,3,10,10,2] and k=3k=3. One of the wonderful colorings of the sequence is shown in the figure.

The example of a wonderful coloring of the sequence a=[3,1,1,1,1,10,3,10,10,2]a=[3,1,1,1,1,10,3,10,10,2] and k=3k=3. Note that one of the elements isn't painted.

Help Paul and Mary to find a wonderful coloring of a given sequence aa.

Input

The first line contains one integer tt (1≤t≤100001≤t≤10000) — the number of test cases. Then tt test cases follow.

Each test case consists of two lines. The first one contains two integers nn and kk (1≤n≤2⋅1051≤n≤2⋅105, 1≤k≤n1≤k≤n) — the length of a given sequence and the number of colors, respectively. The second one contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤n1≤ai≤n).

It is guaranteed that the sum of nn over all test cases doesn't exceed 2⋅1052⋅105.

Output

Output tt 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 nn integers c1,c2,…,cnc1,c2,…,cn (0≤ci≤k0≤ci≤k) separated by spaces where

• ci=0ci=0, if ii-th element isn't painted;
• ci>0ci>0, if ii-th element is painted in the cici-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

Copy

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

Copy

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 11, the blue color — 22, the green — 33.

题目大意

给定一个长度为 nn 的数组 aa，取出 kk 个相离的大小相同的下标集合，集合中每个下标对应的数字需不同，最大化集合的大小，并输出方案。

题目分析

首先统计每个数字出现的次数，如果大于等于 kk 那么显然每个集合分一个，否则待定。贪心地，最后将待定的数字依次 kk 个一组分配给每个集合即可，这里可以用 vector 实现。

#include <bits/stdc++.h>
using namespace std;
int T, n, k, x, ans[200001];
vector<int>num[200001], vec;
int main() {
  cin >> T;
  while (T--) {
    cin >> n;
    cin >> k;
    for ( int i = 1; i <= n; ++i) {
      num[i].clear();
      ans[i] = 0;
    }
    vec.clear();
    for ( int i = 1; i <= n; ++i) {
      cin >> x;
      num[x].push_back(i);
    }
    for ( int i = 1; i <= n; ++i) {
      if (num[i].size() > k)
        for ( int j = 0; j < k; ++j)
          ans[num[i][j]] = j + 1;
      else {
        for ( int j = 0, up = num[i].size(); j < up; ++j)
          vec.push_back(num[i][j]);
      }
    }
    for ( int i = 0, up = vec.size(); i + k - 1 < up; i += k) {
      for ( int j = i; j < i + k; ++j)
        ans[vec[j]] = j - i + 1;
    }
    for ( int i = 1; i <= n; ++i)
      printf("%d ", ans[i]);
    puts("");
  }
  return 0;
}