#include <bits/stdc++.h>
using namespace std;
template <typename T1, typename T2>
inline ostream &operator<<(ostream &os, const pair<T1, T2> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T, size_t N>
inline ostream &operator<<(ostream &os, const array<T, N> &a) {
  os << "[";
  int cnt = 0;
  for (auto &val : a) {
    if (cnt++) os << ", ";
    os << val;
  }
  os << "]";
  return os;
}
template <typename T>
inline ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "[";
  int cnt = 0;
  for (auto &val : v) {
    if (cnt++) os << ", ";
    os << val;
  }
  return os << "]";
}
template <typename T>
inline ostream &operator<<(ostream &os, const set<T> &v) {
  os << "{";
  int cnt = 0;
  for (auto &val : v) {
    if (cnt++) os << ", ";
    os << val;
  }
  return os << "}";
}
template <typename T>
inline ostream &operator<<(ostream &os, const unordered_set<T> &v) {
  os << "{";
  int cnt = 0;
  for (auto &val : v) {
    if (cnt++) os << ", ";
    os << val;
  }
  return os << "}";
}
template <typename T>
inline ostream &operator<<(ostream &os, const multiset<T> &v) {
  os << "{";
  int cnt = 0;
  for (auto &val : v) {
    if (cnt++) os << ", ";
    os << val;
  }
  return os << "}";
}
template <typename T1, typename T2>
inline ostream &operator<<(ostream &os, const map<T1, T2> &v) {
  os << "[";
  int cnt = 0;
  for (auto &val : v) {
    if (cnt++) os << ", ";
    os << val;
  }
  return os << "]";
}
template <typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
  cerr << name << ": " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char *names, Arg1 &&arg1, Args &&...args) {
  const char *comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << ": " << arg1 << " |";
  __f(comma + 1, args...);
}
using ll = long long;
using ull = unsigned long long;
using vi = vector<int>;
using vll = vector<long long>;
using vvi = vector<vector<int>>;
using vs = vector<string>;
using vb = vector<bool>;
using vd = vector<double>;
using vvb = vector<vector<bool>>;
using vvc = vector<vector<char>>;
using vc = vector<char>;
using si = set<int>;
using mpii = map<int, int>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
const int dx[] = {-1, 1, 0, 0};
const int dy[] = {0, 0, 1, -1};
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template <class T>
inline bool umin(T &a, const T b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
inline bool umax(T &a, const T b) {
  return a < b ? a = b, 1 : 0;
}
void solve() {
  int n, k;
  cin >> n >> k;
  vi a(n);
  map<int, vi> pos;
  int rem = 0;
  for (int i = 0; i < (n); i++) {
    cin >> a[i];
    if (((int)(pos[a[i]]).size()) < k) {
      pos[a[i]].push_back(i);
      rem++;
    }
  }
  vi keys;
  for (auto &[c, ps] : pos) {
    keys.emplace_back(c);
  }
  vi b(n, 0);
  int m = ((int)(keys).size());
  int i = 0;
  while (rem >= k) {
    for (int j = 0; j < (k); j++) {
      while (i < m && ((int)(pos[keys[i]]).size()) == 0) i++;
      if (i == m) continue;
      int p = pos[keys[i]].back();
      pos[keys[i]].pop_back();
      b[p] = j + 1;
    }
    rem -= k;
  }
  for (int j = 0; j < (n); j++) cout << b[j] << " \n"[j == n - 1];
}
void solve(int _cas) { solve(); }
int main() {
  srand(time(NULL));
  ios_base::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL), cout.precision(12);
  int cas;
  cin >> cas;
  for (int i = 0; i < (cas); i++) solve(i + 1);
}

### Prompt

Generate 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;
template <typename T1, typename T2>
inline ostream &operator<<(ostream &os, const pair<T1, T2> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T, size_t N>
inline ostream &operator<<(ostream &os, const array<T, N> &a) {
  os << "[";
  int cnt = 0;
  for (auto &val : a) {
    if (cnt++) os << ", ";
    os << val;
  }
  os << "]";
  return os;
}
template <typename T>
inline ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "[";
  int cnt = 0;
  for (auto &val : v) {
    if (cnt++) os << ", ";
    os << val;
  }
  return os << "]";
}
template <typename T>
inline ostream &operator<<(ostream &os, const set<T> &v) {
  os << "{";
  int cnt = 0;
  for (auto &val : v) {
    if (cnt++) os << ", ";
    os << val;
  }
  return os << "}";
}
template <typename T>
inline ostream &operator<<(ostream &os, const unordered_set<T> &v) {
  os << "{";
  int cnt = 0;
  for (auto &val : v) {
    if (cnt++) os << ", ";
    os << val;
  }
  return os << "}";
}
template <typename T>
inline ostream &operator<<(ostream &os, const multiset<T> &v) {
  os << "{";
  int cnt = 0;
  for (auto &val : v) {
    if (cnt++) os << ", ";
    os << val;
  }
  return os << "}";
}
template <typename T1, typename T2>
inline ostream &operator<<(ostream &os, const map<T1, T2> &v) {
  os << "[";
  int cnt = 0;
  for (auto &val : v) {
    if (cnt++) os << ", ";
    os << val;
  }
  return os << "]";
}
template <typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
  cerr << name << ": " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char *names, Arg1 &&arg1, Args &&...args) {
  const char *comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << ": " << arg1 << " |";
  __f(comma + 1, args...);
}
using ll = long long;
using ull = unsigned long long;
using vi = vector<int>;
using vll = vector<long long>;
using vvi = vector<vector<int>>;
using vs = vector<string>;
using vb = vector<bool>;
using vd = vector<double>;
using vvb = vector<vector<bool>>;
using vvc = vector<vector<char>>;
using vc = vector<char>;
using si = set<int>;
using mpii = map<int, int>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
const int dx[] = {-1, 1, 0, 0};
const int dy[] = {0, 0, 1, -1};
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template <class T>
inline bool umin(T &a, const T b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
inline bool umax(T &a, const T b) {
  return a < b ? a = b, 1 : 0;
}
void solve() {
  int n, k;
  cin >> n >> k;
  vi a(n);
  map<int, vi> pos;
  int rem = 0;
  for (int i = 0; i < (n); i++) {
    cin >> a[i];
    if (((int)(pos[a[i]]).size()) < k) {
      pos[a[i]].push_back(i);
      rem++;
    }
  }
  vi keys;
  for (auto &[c, ps] : pos) {
    keys.emplace_back(c);
  }
  vi b(n, 0);
  int m = ((int)(keys).size());
  int i = 0;
  while (rem >= k) {
    for (int j = 0; j < (k); j++) {
      while (i < m && ((int)(pos[keys[i]]).size()) == 0) i++;
      if (i == m) continue;
      int p = pos[keys[i]].back();
      pos[keys[i]].pop_back();
      b[p] = j + 1;
    }
    rem -= k;
  }
  for (int j = 0; j < (n); j++) cout << b[j] << " \n"[j == n - 1];
}
void solve(int _cas) { solve(); }
int main() {
  srand(time(NULL));
  ios_base::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL), cout.precision(12);
  int cas;
  cin >> cas;
  for (int i = 0; i < (cas); i++) solve(i + 1);
}

```