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.
12
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
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
 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

2 1 2 3 0 1 3 2 3 1 
1 2 3 4 
1 
1 1 1 0 1 1 1 1 0 0 0 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 
2 1 2 3 0 1 3 2 3 1 
1 2 3 4 
1 
1 1 1 0 1 1 1 1 0 0 0 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 
 2 1 2 3 0 1 3 2 3 1 
1 2 3 4 
1 
1 1 1 0 1 1 1 1 0 0 0 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 

#include <bits/stdc++.h>
#pragma comment(linker, "/stack:16777216")
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
const long long MOD = 1e9 + 7;
const long long LOG = 19;
const long long INF = 1e18;
const double PI = acos(-1);
const double EPS = 1e-9;
const long long N = 3e5 + 5;
long long n, k;
vector<long long> a;
map<long long, vector<long long> > q;
bool comp(pair<long long, vector<long long> > a,
          pair<long long, vector<long long> > b) {
  return a.second.size() > b.second.size();
}
bool check(long long x, vector<pair<long long, vector<long long> > > q) {
  vector<pair<long long, vector<long long> > > cnt = q;
  for (long long i = 0; i < k; i++) {
    set<long long> st;
    for (long long j = 0; j < cnt.size(); j++) {
      if (st.size() == x) break;
      if (!st.count(cnt[j].first) && cnt[j].second.size()) {
        st.insert(cnt[j].first);
        cnt[j].second.pop_back();
      }
    }
    if (st.size() != x) return false;
  }
  return true;
}
void solve() {
  cin >> n >> k;
  a.resize(n);
  q.clear();
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    if (q[a[i]].size() < k) q[a[i]].push_back(i);
  }
  vector<long long> res(n, 0);
  vector<pair<long long, vector<long long> > > cnt;
  for (auto x : q) {
    cnt.push_back(x);
  }
  sort(cnt.begin(), cnt.end(), comp);
  long long l = 1, r = n;
  for (long long j = 0; j < 20; j++) {
    long long mid = (l + r) / 2;
    if (check(mid, cnt)) {
      l = mid;
    } else {
      r = mid;
    }
  }
  if (check(r, cnt)) {
    for (long long i = 1; i <= k; i++) {
      set<long long> st;
      for (long long j = 0; j < cnt.size(); j++) {
        if (st.size() == r) break;
        if (!st.count(cnt[j].first) && cnt[j].second.size()) {
          st.insert(cnt[j].first);
          res[cnt[j].second.back()] = i;
          cnt[j].second.pop_back();
        }
      }
    }
    for (long long i : res) cout << i << ' ';
  } else {
    for (long long i = 1; i <= k; i++) {
      set<long long> st;
      for (long long j = 0; j < cnt.size(); j++) {
        if (st.size() == l) break;
        if (!st.count(cnt[j].first) && cnt[j].second.size()) {
          st.insert(cnt[j].first);
          res[cnt[j].second.back()] = i;
          cnt[j].second.pop_back();
        }
      }
    }
    for (long long i : res) cout << i << ' ';
  }
  cout << endl;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}
