#include <bits/stdc++.h>
using namespace std;
long long power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = res * x;
    y = y >> 1;
    x = x * x;
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long pow(long long a, long long b, long long MOD) {
  long long x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x % MOD * y % MOD);
      if (x > MOD) x %= MOD;
    }
    y = (y % MOD * y % MOD);
    if (y > MOD) y %= MOD;
    b /= 2;
  }
  return x;
}
void union1(long long parent[], int x, int y) { parent[y] = x; }
int find1(long long parent[], int x) {
  if (parent[x] == -1) {
    return x;
  }
  return find1(parent, parent[x]);
}
int main(void) {
  long long t, n, x, y, k, m, r, p, d, u, v, q, a, b, z, w, l;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    unordered_map<long long, long long> mp;
    unordered_map<long long, vector<long long>> v1;
    unordered_map<long long, long long> indices;
    vector<long long> v(n);
    for (int i = 0; i < n; i++) {
      cin >> v[i];
      mp[v[i]]++;
    }
    vector<pair<int, int>> p;
    long long res = 0;
    for (auto x : mp) {
      if (x.second >= k) {
        long long r = k, c = 1;
        while (r--) {
          v1[x.first].push_back(c++);
        }
      } else {
        p.push_back({x.second, x.first});
        res += x.second;
      }
    }
    int f = 0, c1 = 0;
    long long temp = res % k;
    res -= temp;
    for (int i = 0; i < p.size(); i++) {
      while (c1 < res && p[i].first--) {
        c1++;
        v1[p[i].second].push_back((f % k) + 1);
        f++;
      }
    }
    for (auto x : v1) {
      indices[x.first] = 0;
    }
    for (int i = 0; i < n; i++) {
      if (indices[v[i]] < v1[v[i]].size()) {
        cout << v1[v[i]][indices[v[i]]++] << " ";
      } else
        cout << 0 << " ";
    }
    cout << endl;
  }
}

### Prompt

In cpp, your task is to solve 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;
long long power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = res * x;
    y = y >> 1;
    x = x * x;
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long pow(long long a, long long b, long long MOD) {
  long long x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x % MOD * y % MOD);
      if (x > MOD) x %= MOD;
    }
    y = (y % MOD * y % MOD);
    if (y > MOD) y %= MOD;
    b /= 2;
  }
  return x;
}
void union1(long long parent[], int x, int y) { parent[y] = x; }
int find1(long long parent[], int x) {
  if (parent[x] == -1) {
    return x;
  }
  return find1(parent, parent[x]);
}
int main(void) {
  long long t, n, x, y, k, m, r, p, d, u, v, q, a, b, z, w, l;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    unordered_map<long long, long long> mp;
    unordered_map<long long, vector<long long>> v1;
    unordered_map<long long, long long> indices;
    vector<long long> v(n);
    for (int i = 0; i < n; i++) {
      cin >> v[i];
      mp[v[i]]++;
    }
    vector<pair<int, int>> p;
    long long res = 0;
    for (auto x : mp) {
      if (x.second >= k) {
        long long r = k, c = 1;
        while (r--) {
          v1[x.first].push_back(c++);
        }
      } else {
        p.push_back({x.second, x.first});
        res += x.second;
      }
    }
    int f = 0, c1 = 0;
    long long temp = res % k;
    res -= temp;
    for (int i = 0; i < p.size(); i++) {
      while (c1 < res && p[i].first--) {
        c1++;
        v1[p[i].second].push_back((f % k) + 1);
        f++;
      }
    }
    for (auto x : v1) {
      indices[x.first] = 0;
    }
    for (int i = 0; i < n; i++) {
      if (indices[v[i]] < v1[v[i]].size()) {
        cout << v1[v[i]][indices[v[i]]++] << " ";
      } else
        cout << 0 << " ";
    }
    cout << endl;
  }
}

```