#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const long long mxn = 1e5 + 7;
const long long inf = 1e18;
inline void normal(long long &a) {
  a %= mod;
  (a < 0) && (a += mod);
}
inline long long modMul(long long a, long long b) {
  a %= mod, b %= mod;
  normal(a), normal(b);
  return (a * b) % mod;
}
inline long long modAdd(long long a, long long b) {
  a %= mod, b %= mod;
  normal(a), normal(b);
  return (a + b) % mod;
}
inline long long modSub(long long a, long long b) {
  a %= mod, b %= mod;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long modPow(long long b, long long p) {
  long long r = 1;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1;
  }
  return r;
}
inline long long modInverse(long long a) { return modPow(a, mod - 2); }
inline long long modDiv(long long a, long long b) {
  return modMul(a, modInverse(b));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long i, j, k, a, b, c, d, n, m, t, u, v, w, x, y, z;
  cin >> t;
  for (long long zz = 1; zz <= t; zz++) {
    cin >> n >> k;
    vector<long long> vec(n);
    set<long long> st;
    map<long long, long long> frq;
    map<long long, vector<long long> > pos;
    for (i = 0; i < n; i++) {
      cin >> vec[i];
      a = vec[i];
      st.insert(a);
      frq[a] += 1;
      pos[a].push_back(i);
    }
    vector<long long> ans(n, 0), tem1, tem2;
    for (auto it : st) {
      if (frq[it] >= k) {
        tem1.push_back(it);
      } else {
        tem2.push_back(it);
      }
    }
    for (auto a : tem1) {
      long long col = 0;
      for (auto b : pos[a]) {
        col += 1;
        ans[b] = col;
        if (col == k) {
          break;
        }
      }
    }
    long long col = 0, cnt = (long long)tem1.size();
    for (auto a : tem2) {
      for (auto b : pos[a]) {
        long long tem = (col % k) + 1;
        ans[b] = tem;
        if (tem == k) {
          cnt += 1;
        }
        col += 1;
      }
    }
    map<long long, long long> check;
    for (i = 0; i < n; i++) {
      a = ans[i];
      if (a >= 1) {
        check[a] += 1;
      }
      if (check[a] <= cnt) {
        cout << ans[i] << " ";
      } else {
        cout << 0 << " ";
      }
    }
    cout << "\n";
  }
  return 0;
}

### Prompt

Please formulate 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;
const long long mod = 1e9 + 7;
const long long mxn = 1e5 + 7;
const long long inf = 1e18;
inline void normal(long long &a) {
  a %= mod;
  (a < 0) && (a += mod);
}
inline long long modMul(long long a, long long b) {
  a %= mod, b %= mod;
  normal(a), normal(b);
  return (a * b) % mod;
}
inline long long modAdd(long long a, long long b) {
  a %= mod, b %= mod;
  normal(a), normal(b);
  return (a + b) % mod;
}
inline long long modSub(long long a, long long b) {
  a %= mod, b %= mod;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long modPow(long long b, long long p) {
  long long r = 1;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1;
  }
  return r;
}
inline long long modInverse(long long a) { return modPow(a, mod - 2); }
inline long long modDiv(long long a, long long b) {
  return modMul(a, modInverse(b));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long i, j, k, a, b, c, d, n, m, t, u, v, w, x, y, z;
  cin >> t;
  for (long long zz = 1; zz <= t; zz++) {
    cin >> n >> k;
    vector<long long> vec(n);
    set<long long> st;
    map<long long, long long> frq;
    map<long long, vector<long long> > pos;
    for (i = 0; i < n; i++) {
      cin >> vec[i];
      a = vec[i];
      st.insert(a);
      frq[a] += 1;
      pos[a].push_back(i);
    }
    vector<long long> ans(n, 0), tem1, tem2;
    for (auto it : st) {
      if (frq[it] >= k) {
        tem1.push_back(it);
      } else {
        tem2.push_back(it);
      }
    }
    for (auto a : tem1) {
      long long col = 0;
      for (auto b : pos[a]) {
        col += 1;
        ans[b] = col;
        if (col == k) {
          break;
        }
      }
    }
    long long col = 0, cnt = (long long)tem1.size();
    for (auto a : tem2) {
      for (auto b : pos[a]) {
        long long tem = (col % k) + 1;
        ans[b] = tem;
        if (tem == k) {
          cnt += 1;
        }
        col += 1;
      }
    }
    map<long long, long long> check;
    for (i = 0; i < n; i++) {
      a = ans[i];
      if (a >= 1) {
        check[a] += 1;
      }
      if (check[a] <= cnt) {
        cout << ans[i] << " ";
      } else {
        cout << 0 << " ";
      }
    }
    cout << "\n";
  }
  return 0;
}

```