#include <bits/stdc++.h>
/*Author: _KitCat_*/
#pragma GCC optimize ("O3")

#define ll long long
#define pii pair<long long, long long>
#define vi vector<long long>
#define pb push_back
#define mp make_pair
#define um unordered_map
#define iospeed cin.tie(0);cout.tie(0);ios_base::sync_with_stdio(0);
#define fin for(int i = 0; i < n; i++)
#define rep(a,b,c) for(int a = b; a < c; a++)
#define TESTS int tests; cin >> tests; while(tests--)
#define nl '\n'
#define print(STRUCT) for(auto VALUE : STRUCT) cout << VALUE << ' '; cout << endl;
#define all(x) x.begin(), x.end()
#define int long long
#define f first
#define s second

using namespace std;
struct Comparator{
    bool operator() (const pair<int, int>& lhs,const pair<int, int>& rhs) const{
      if (lhs.first == rhs.first) return lhs.second < rhs.second;
      return lhs.first < rhs.first;
  }
    bool operator() (int a, int b) const {return a > b;}      
};

void solve(){
    int n;
    cin >> n;
    vi arr(n);
    map<int, int> count; //remember how many times each element occurs 
    fin {  
        cin >> arr[i];
        ++count[arr[i]];
    }
    vi x(n);
    vi y(n);
    set<int> S; //set keeps distinct elements that occur to the left of 
    //current index and won`t be futhermore 

    fin{
        x[i] = S.size();
        --count[arr[i]]; 
        if(count[arr[i]] == 0) S.insert(arr[i]);
        //if there would be no more such elements to the right of current index
        //we can add it to our set
    }
    //resetting set and map
    S.clear();
    count.clear();
    fin ++count[arr[i]];

    for(int i = n-1; i >= 0; i--){
        y[i] = S.size();
        --count[arr[i]];
        if(count[arr[i]] == 0)
            S.insert(arr[i]);
    }
    fin{
        cout << abs(x[i]-y[i]) << ' ';
    }
}

signed main() {
  iospeed;
  TESTS
    solve();
  

  return 0;
}