B2. Wonderful Coloring - 2

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 a1,a2,…,ana1,a2,…,an. They want to paint it using pieces of chalk of kk 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 kk 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 kk 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]a=[3,1,1,1,1,10,3,10,10,2] and k=3k=3. One of the wonderful colorings of the sequence is shown in the figure.

The example of a wonderful coloring of the sequence a=[3,1,1,1,1,10,3,10,10,2]a=[3,1,1,1,1,10,3,10,10,2] and k=3k=3. Note that one of the elements isn't painted.

Help Paul and Mary to find a wonderful coloring of a given sequence aa.

Input

The first line contains one integer tt (1≤t≤100001≤t≤10000) — the number of test cases. Then tt test cases follow.

Each test case consists of two lines. The first one contains two integers nn and kk (1≤n≤2⋅1051≤n≤2⋅105, 1≤k≤n1≤k≤n) — the length of a given sequence and the number of colors, respectively. The second one contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤n1≤ai≤n).

It is guaranteed that the sum of nn over all test cases doesn't exceed 2⋅1052⋅105.

Output

Output tt 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 nn integers c1,c2,…,cnc1,c2,…,cn (0≤ci≤k0≤ci≤k) separated by spaces where

• ci=0ci=0, if ii-th element isn't painted;
• ci>0ci>0, if ii-th element is painted in the cici-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

Copy

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

Copy

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 11, the blue color — 22, the green — 33.

 代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
using namespace std;
typedef long long ll;
const int maxn=5e5+100;
int vis[maxn];
int pp[maxn];
int ans[maxn];
vector<int>v[maxn];
struct node{
    int w;
    int id;
}a[maxn]; 
bool cmp(node x,node y){
    return x.w<y.w;
}
int main(){
    int t;
    cin>>t;
    while(t--){
        memset(pp,0,sizeof(pp));
        memset(ans,0,sizeof(ans)); 
        int n,k;
        cin>>n>>k;
        for(int i=1;i<=n;i++){
            v[i].clear();
            cin>>a[i].w;
            //vis[a[i].w]++;
            a[i].id=i;
        } 
        sort(a+1,a+n+1,cmp);
        int cnt=1;
        for(int i=1;i<=n;i++){
            if(v[cnt].size()==k){
                cnt++;
            }
            if(pp[a[i].w]<k){
                v[cnt].push_back(a[i].id);
                pp[a[i].w]++;
            }
        }
        for(int i=0;i<=cnt;i++){
            if(v[i].size()<k){
                continue;
            }
            else{
                for(int j=0;j<v[i].size();j++){
                    ans[v[i][j]]=j+1;
                }
            }
        }
        for(int i=1;i<=n;i++){
            cout<<ans[i]<<" ";
        }
        cout<<endl;
    }
}