#include <bits/stdc++.h>
using ll = long long;
using namespace std;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
bool isPrime(ll n) {
  if (n <= 1) return false;
  for (ll i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
void sieve(ll n) {
  vector<bool> prime(n + 1, true);
  for (ll p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (ll i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
}
ll gcd(ll a, ll b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
ll modular(ll x, unsigned long long int y, ll p) {
  ll res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
int main() {
  ll t;
  cin >> t;
  while (t--) {
    ll n;
    cin >> n;
    ll k;
    cin >> k;
    vector<ll> v(n);
    ll i;
    map<ll, vector<ll>> m;
    vector<ll> ans(n, 0LL);
    for (i = 0; i < n; i++) {
      cin >> v[i];
      if (m[v[i]].size() < k) m[v[i]].push_back(i);
    };
    ll sz = 0;
    ll z = 0;
    ll tt = 0;
    for (auto i : m) z += i.second.size();
    z -= z % k;
    for (auto i = m.begin(); i != m.end(); i++) {
      if (z == 0) break;
      for (ll j = 0; j < min((ll)i->second.size(), k); j++) {
        ans[i->second[j]] = ++tt;
        tt = tt % k;
        z--;
        if (z == 0) break;
      }
    }
    for (auto i : ans) cout << i << " ";
    cout << "\n";
  }
  return 0;
}

### Prompt

Your challenge is to write 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 ll = long long;
using namespace std;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
bool isPrime(ll n) {
  if (n <= 1) return false;
  for (ll i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
void sieve(ll n) {
  vector<bool> prime(n + 1, true);
  for (ll p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (ll i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
}
ll gcd(ll a, ll b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
ll modular(ll x, unsigned long long int y, ll p) {
  ll res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
int main() {
  ll t;
  cin >> t;
  while (t--) {
    ll n;
    cin >> n;
    ll k;
    cin >> k;
    vector<ll> v(n);
    ll i;
    map<ll, vector<ll>> m;
    vector<ll> ans(n, 0LL);
    for (i = 0; i < n; i++) {
      cin >> v[i];
      if (m[v[i]].size() < k) m[v[i]].push_back(i);
    };
    ll sz = 0;
    ll z = 0;
    ll tt = 0;
    for (auto i : m) z += i.second.size();
    z -= z % k;
    for (auto i = m.begin(); i != m.end(); i++) {
      if (z == 0) break;
      for (ll j = 0; j < min((ll)i->second.size(), k); j++) {
        ans[i->second[j]] = ++tt;
        tt = tt % k;
        z--;
        if (z == 0) break;
      }
    }
    for (auto i : ans) cout << i << " ";
    cout << "\n";
  }
  return 0;
}

```