#include <bits/stdc++.h>
using namespace std;
using ll = long long;
template <class T, class U>
inline bool chmax(T& a, U b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T, class U>
inline bool chmin(T& a, U b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
constexpr int INF = 1000000000;
constexpr ll llINF = 3000000000000000000;
constexpr int mod = 1000000007;
constexpr double eps = 1e-10;
const double pi = acos(-1);
int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};
int Random(int mi, int ma) {
  random_device rnd;
  mt19937 mt(rnd());
  uniform_int_distribution<int> engine(mi, ma);
  return engine(mt);
}
bool prime(int a) {
  if (a == 1) return false;
  for (int i = 2; i * i <= a; i++) {
    if (a % i == 0) return false;
  }
  return true;
}
vector<int> primes;
void init_prime(int n) {
  primes.push_back(2);
  for (int i = 3; i <= n; i += 2) {
    bool f = true;
    for (int j : primes) {
      if (j * j > i) break;
      if (i % j == 0) {
        f = false;
        break;
      }
    }
    if (f) primes.push_back(i);
  }
}
ll modpow(ll a, ll b) {
  ll res = 1;
  while (b) {
    if (b & 1) {
      res *= a;
      res %= mod;
    }
    a *= a;
    a %= mod;
    b >>= 1;
  }
  return res;
}
vector<ll> inv, fact, factinv;
void init_fact(int n) {
  inv.resize(n + 1);
  fact.resize(n + 1);
  factinv.resize(n + 1);
  inv[0] = 0;
  inv[1] = 1;
  fact[0] = 1;
  factinv[0] = 1;
  for (ll i = 1; i <= n; i++) {
    if (i >= 2) inv[i] = mod - ((mod / i) * inv[mod % i] % mod);
    fact[i] = (fact[i - 1] * i) % mod;
    factinv[i] = factinv[i - 1] * inv[i] % mod;
  }
}
ll _inv(ll a, ll m = mod) {
  ll b = m, u = 1, v = 0;
  while (b) {
    ll t = a / b;
    a -= t * b;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  u %= m;
  if (u < 0) u += m;
  return u;
}
ll comb(int a, int b) {
  if (a < b) return 0;
  if (a < 0) return 0;
  return fact[a] * factinv[a - b] % mod * factinv[b] % mod;
}
void solve() {
  int n, k;
  cin >> n >> k;
  vector<vector<int>> v(n);
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    a--;
    v[a].push_back(i);
  }
  vector<int> ans(n);
  int s = 0;
  for (int i = 0; i < n; i++) s += min(k, (int)v[i].size());
  s = s / k * k;
  int cur = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < min(k, (int)v[i].size()); j++) {
      if (cur < s) {
        ans[v[i][j]] = (cur % k) + 1;
        cur++;
      }
    }
  }
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
  cout << endl;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) solve();
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
using ll = long long;
template <class T, class U>
inline bool chmax(T& a, U b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T, class U>
inline bool chmin(T& a, U b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
constexpr int INF = 1000000000;
constexpr ll llINF = 3000000000000000000;
constexpr int mod = 1000000007;
constexpr double eps = 1e-10;
const double pi = acos(-1);
int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};
int Random(int mi, int ma) {
  random_device rnd;
  mt19937 mt(rnd());
  uniform_int_distribution<int> engine(mi, ma);
  return engine(mt);
}
bool prime(int a) {
  if (a == 1) return false;
  for (int i = 2; i * i <= a; i++) {
    if (a % i == 0) return false;
  }
  return true;
}
vector<int> primes;
void init_prime(int n) {
  primes.push_back(2);
  for (int i = 3; i <= n; i += 2) {
    bool f = true;
    for (int j : primes) {
      if (j * j > i) break;
      if (i % j == 0) {
        f = false;
        break;
      }
    }
    if (f) primes.push_back(i);
  }
}
ll modpow(ll a, ll b) {
  ll res = 1;
  while (b) {
    if (b & 1) {
      res *= a;
      res %= mod;
    }
    a *= a;
    a %= mod;
    b >>= 1;
  }
  return res;
}
vector<ll> inv, fact, factinv;
void init_fact(int n) {
  inv.resize(n + 1);
  fact.resize(n + 1);
  factinv.resize(n + 1);
  inv[0] = 0;
  inv[1] = 1;
  fact[0] = 1;
  factinv[0] = 1;
  for (ll i = 1; i <= n; i++) {
    if (i >= 2) inv[i] = mod - ((mod / i) * inv[mod % i] % mod);
    fact[i] = (fact[i - 1] * i) % mod;
    factinv[i] = factinv[i - 1] * inv[i] % mod;
  }
}
ll _inv(ll a, ll m = mod) {
  ll b = m, u = 1, v = 0;
  while (b) {
    ll t = a / b;
    a -= t * b;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  u %= m;
  if (u < 0) u += m;
  return u;
}
ll comb(int a, int b) {
  if (a < b) return 0;
  if (a < 0) return 0;
  return fact[a] * factinv[a - b] % mod * factinv[b] % mod;
}
void solve() {
  int n, k;
  cin >> n >> k;
  vector<vector<int>> v(n);
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    a--;
    v[a].push_back(i);
  }
  vector<int> ans(n);
  int s = 0;
  for (int i = 0; i < n; i++) s += min(k, (int)v[i].size());
  s = s / k * k;
  int cur = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < min(k, (int)v[i].size()); j++) {
      if (cur < s) {
        ans[v[i][j]] = (cur % k) + 1;
        cur++;
      }
    }
  }
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
  cout << endl;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) solve();
}

```