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.
12
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
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
 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

2 1 2 3 0 1 3 2 3 1 
1 2 3 4 
1 
1 1 1 0 1 1 1 1 0 0 0 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 
2 1 2 3 0 1 3 2 3 1 
1 2 3 4 
1 
1 1 1 0 1 1 1 1 0 0 0 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 
 2 1 2 3 0 1 3 2 3 1 
1 2 3 4 
1 
1 1 1 0 1 1 1 1 0 0 0 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 

#include <bits/stdc++.h>

#define sz(x) int(x.size())
#define uni(x) x.resize(unique(x.begin(), x.end()) - x.begin())
#define ms(x, n) memset(x, n, sizeof(x))
#define mc(a, b) memcpy(a, b, sizeof(b))
#define tomax(a, b) a = max(a, b)
#define tomin(a, b) a = min(a, b)

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<int, ll> pil;
typedef pair<ll, int> pli;
typedef pair<ll, ll> pll;

const int inf = 0x3f3f3f3f;
const int ninf = 0xc0c0c0c0;
const ll linf = 0x3f3f3f3f3f3f3f3fll;
const ll lninf = 0xc0c0c0c0c0c0c0c0ll;

template<typename T, typename = void> struct is_container : false_type {};
template<typename T> struct is_container<T, void_t<typename T::iterator>> : true_type {};
template<> struct is_container<string> : false_type {};
template<typename T, typename = void> struct is_tuple : false_type {};
template<typename T> struct is_tuple<T, void_t<typename tuple_size<T>::type>> : true_type {};

template<typename Ch, typename Tr, typename T>
basic_ostream<Ch, Tr>& operator<<(basic_ostream<Ch, Tr>& os, const T& obj) {
    if constexpr (is_container<T>::value) {
        int n = 0;
        const char* sep = is_container<typename T::value_type>::value || is_tuple<typename T::value_type>::value ? "\n" : " ";
        for (const auto& val : obj) os << val << (++n < obj.size() ? sep : "");
    }
    else if constexpr (is_tuple<T>::value) {
        apply([&os](const auto&... args) {
            int n = 0;
            ((os << args << (++n < sizeof...(args) ? " " : "")), ...);
        }, obj);
    }
    return os;
}

template<typename... Args>
void print(const Args&... args) {
    int n = 0;
    ((cout << args << (++n < sizeof...(args) ? " " : "")), ...);
    cout << endl;
}

int ans[200005];
pii a[200005];

void run() {
    int m,n;
    cin>>n>>m;
    for(int i=1;i<=n;i++) {
        cin>>a[i].first;
        a[i].second=i;
        ans[i]=0;
    }
    sort(a+1,a+1+n);
    int x=1;
    for(int i=1;i<=n;i++) {
        ans[a[i].second]=x;
        x=x%m+1;
        if(i-m+1>=1&&a[i].first==a[i-m+1].first) {
            while(i+1<=n&&a[i+1].first==a[i].first) i++;
        }
        if(x==1&&n-i<m) break;
    }
    for(int i=1;i<=n;i++) cout<<ans[i]<<' ';
    cout<<endl;
}

int main() {
    ios::sync_with_stdio(false);
    int T;
    cin >> T;
    for (int I = 1; I <= T; I++) run();
    return 0;
}
