#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
using namespace std;
long long MOD = 998244353;
long long mod = 1000000007;
double eps = 1e-12;
double pi = acos(-1);
mt19937 RNG(chrono::steady_clock::now().time_since_epoch().count());
template <typename T>
T gcd(T a, T b) {
  return (b ? __gcd(a, b) : a);
}
template <typename T>
T lcm(T a, T b) {
  return (a * (b / gcd(a, b)));
}
int add(int a, int b, int c = MOD) {
  int res = a + b;
  return (res >= c ? res - c : res);
}
int mod_neg(int a, int b, int c = MOD) {
  int res;
  if (abs(a - b) < c)
    res = a - b;
  else
    res = (a - b) % c;
  return (res < 0 ? res + c : res);
}
int mul(int a, int b, int c = MOD) {
  long long res = (long long)a * b;
  return (res >= c ? res % c : res);
}
int muln(int a, int b, int c = MOD) {
  long long res = (long long)a * b;
  return ((res % c) + c) % c;
}
long long mulmod(long long a, long long b, long long m = MOD) {
  long long q = (long long)(((long double)a * (long double)b) / (long double)m);
  long long r = a * b - q * m;
  if (r > m) r %= m;
  if (r < 0) r += m;
  return r;
}
template <typename T>
T expo(T e, T n) {
  T x = 1, p = e;
  while (n) {
    if (n & 1) x = x * p;
    p = p * p;
    n >>= 1;
  }
  return x;
}
template <typename T>
T power(T e, T n, T m = MOD) {
  T x = 1, p = e;
  while (n) {
    if (n & 1) x = mul(x, p, m);
    p = mul(p, p, m);
    n >>= 1;
  }
  return x;
}
template <typename T>
T extended_euclid(T a, T b, T &x, T &y) {
  T xx = 0, yy = 1;
  y = 0;
  x = 1;
  while (b) {
    T q = a / b, t = b;
    b = a % b;
    a = t;
    t = xx;
    xx = x - q * xx;
    x = t;
    t = yy;
    yy = y - q * yy;
    y = t;
  }
  return a;
}
template <typename T>
T mod_inverse(T a, T n = MOD) {
  T x, y, z = 0;
  T d = extended_euclid(a, n, x, y);
  return (d > 1 ? -1 : mod_neg(x, z, n));
}
const int FACSZ = 1;
int fact[FACSZ], ifact[FACSZ];
void precom(int c = MOD) {
  fact[0] = 1;
  for (long long i = 1; i < FACSZ; i++) fact[i] = mul(fact[i - 1], i, c);
  ifact[FACSZ - 1] = mod_inverse(fact[FACSZ - 1], c);
  for (long long i = 0; i < FACSZ - 1; i++) {
    ifact[i] = mul(i + 1, ifact[i + 1], c);
  }
}
int ncr(int n, int r, int c = MOD) {
  return mul(mul(ifact[r], ifact[n - r], c), fact[n], c);
}
void solve() {
  long long n, k;
  cin >> n >> k;
  long long a[n];
  map<int, vector<int> > m;
  vector<long long> ans(n, 0);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    m[a[i]].push_back(i);
  }
  long long count = 0;
  long long count2 = 0;
  for (auto x : m) {
    if (x.second.size() < k) {
      count += x.second.size();
    } else {
      count2++;
      for (long long i = 0; i < k; i++) {
        ans[x.second[i]] = i + 1;
      }
    }
  }
  long long g = 0;
  for (auto x : m) {
    if (x.second.size() < k) {
      for (long long i = 0; i < x.second.size(); i++) {
        if (g < count - count % k) {
          ans[x.second[i]] = g % k + 1;
          g++;
        }
      }
    }
  }
  for (long long i = 0; i < n; i++) {
    cout << ans[i] << " ";
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  for (int it = 1; it <= t; it++) {
    solve();
  }
  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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
using namespace std;
long long MOD = 998244353;
long long mod = 1000000007;
double eps = 1e-12;
double pi = acos(-1);
mt19937 RNG(chrono::steady_clock::now().time_since_epoch().count());
template <typename T>
T gcd(T a, T b) {
  return (b ? __gcd(a, b) : a);
}
template <typename T>
T lcm(T a, T b) {
  return (a * (b / gcd(a, b)));
}
int add(int a, int b, int c = MOD) {
  int res = a + b;
  return (res >= c ? res - c : res);
}
int mod_neg(int a, int b, int c = MOD) {
  int res;
  if (abs(a - b) < c)
    res = a - b;
  else
    res = (a - b) % c;
  return (res < 0 ? res + c : res);
}
int mul(int a, int b, int c = MOD) {
  long long res = (long long)a * b;
  return (res >= c ? res % c : res);
}
int muln(int a, int b, int c = MOD) {
  long long res = (long long)a * b;
  return ((res % c) + c) % c;
}
long long mulmod(long long a, long long b, long long m = MOD) {
  long long q = (long long)(((long double)a * (long double)b) / (long double)m);
  long long r = a * b - q * m;
  if (r > m) r %= m;
  if (r < 0) r += m;
  return r;
}
template <typename T>
T expo(T e, T n) {
  T x = 1, p = e;
  while (n) {
    if (n & 1) x = x * p;
    p = p * p;
    n >>= 1;
  }
  return x;
}
template <typename T>
T power(T e, T n, T m = MOD) {
  T x = 1, p = e;
  while (n) {
    if (n & 1) x = mul(x, p, m);
    p = mul(p, p, m);
    n >>= 1;
  }
  return x;
}
template <typename T>
T extended_euclid(T a, T b, T &x, T &y) {
  T xx = 0, yy = 1;
  y = 0;
  x = 1;
  while (b) {
    T q = a / b, t = b;
    b = a % b;
    a = t;
    t = xx;
    xx = x - q * xx;
    x = t;
    t = yy;
    yy = y - q * yy;
    y = t;
  }
  return a;
}
template <typename T>
T mod_inverse(T a, T n = MOD) {
  T x, y, z = 0;
  T d = extended_euclid(a, n, x, y);
  return (d > 1 ? -1 : mod_neg(x, z, n));
}
const int FACSZ = 1;
int fact[FACSZ], ifact[FACSZ];
void precom(int c = MOD) {
  fact[0] = 1;
  for (long long i = 1; i < FACSZ; i++) fact[i] = mul(fact[i - 1], i, c);
  ifact[FACSZ - 1] = mod_inverse(fact[FACSZ - 1], c);
  for (long long i = 0; i < FACSZ - 1; i++) {
    ifact[i] = mul(i + 1, ifact[i + 1], c);
  }
}
int ncr(int n, int r, int c = MOD) {
  return mul(mul(ifact[r], ifact[n - r], c), fact[n], c);
}
void solve() {
  long long n, k;
  cin >> n >> k;
  long long a[n];
  map<int, vector<int> > m;
  vector<long long> ans(n, 0);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    m[a[i]].push_back(i);
  }
  long long count = 0;
  long long count2 = 0;
  for (auto x : m) {
    if (x.second.size() < k) {
      count += x.second.size();
    } else {
      count2++;
      for (long long i = 0; i < k; i++) {
        ans[x.second[i]] = i + 1;
      }
    }
  }
  long long g = 0;
  for (auto x : m) {
    if (x.second.size() < k) {
      for (long long i = 0; i < x.second.size(); i++) {
        if (g < count - count % k) {
          ans[x.second[i]] = g % k + 1;
          g++;
        }
      }
    }
  }
  for (long long i = 0; i < n; i++) {
    cout << ans[i] << " ";
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  for (int it = 1; it <= t; it++) {
    solve();
  }
  return 0;
}

```