description: Dmitry has an array of n non-negative integers a_1, a_2, ..., a_n.
  
  In one operation, Dmitry can choose any index j (1 ≤ j ≤ n) and increase the value of the element a_j by 1. He can choose the same index j multiple times.
  
  For each i from 0 to n, determine whether Dmitry can make the MEX of the array equal to exactly i. If it is possible, then determine the minimum number of operations to do it.
  
  The MEX of the array is equal to the minimum non-negative integer that is not in the array. For example, the MEX of the array [3, 1, 0] is equal to 2, and the array [3, 3, 1, 4] is equal to 0.
  
  Input
  
  The first line of input data contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases in the input. 
  
  The descriptions of the test cases follow.
  
  The first line of the description of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the array a.
  
  The second line of the description of each test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ n) — elements of the array a.
  
  It is guaranteed that the sum of the values n over all test cases in the test does not exceed 2⋅10^5.
  
  Output
  
  For each test case, output n + 1 integer — i-th number is equal to the minimum number of operations for which you can make the array MEX equal to i (0 ≤ i ≤ n), or -1 if this cannot be done.
  
  Example
  
  Input
  
  
  5
  3
  0 1 3
  7
  0 1 2 3 4 3 2
  4
  3 0 0 0
  7
  4 6 2 3 5 0 5
  5
  4 0 1 0 4
  
  
  Output
  
  
  1 1 0 -1 
  1 1 2 2 1 0 2 6 
  3 0 1 4 3 
  1 0 -1 -1 -1 -1 -1 -1 
  2 1 0 2 -1 -1 
  
  Note
  
  In the first set of example inputs, n=3:
  
    * to get MEX=0, it is enough to perform one increment: a_1++; 
    * to get MEX=1, it is enough to perform one increment: a_2++; 
    * MEX=2 for a given array, so there is no need to perform increments; 
    * it is impossible to get MEX=3 by performing increments. 
solution:  //package eround101;
  
  import java.util.*;
  import java.io.*;
  import java.lang.*;
  import java.util.StringTokenizer;
  
  public class Solution {
  
      static HritikScanner sc = new HritikScanner();
      static PrintWriter pw = new PrintWriter(System.out, true);
      final static int MOD = 1000000007;
  
      public static void main(String[] args) {
  
          int t = ni();
          while (t-- > 0) {
              solve();
          }
      }
  
      static void solve() {
          int n = ni();
          int[] arr = nextIntArray(n);
          Arrays.sort(arr);
          Map<Integer, Integer> map = new HashMap<>();
          for(int i = 0; i< n; i++)
          {
              map.put(arr[i], map.getOrDefault(arr[i], 0)+1);
          }
          Stack<Integer> stk = new Stack<>();
          long[] dp = new long[n+1];
          long[] ans = new long[n+1];
          //Arrays.fill(ans, -1);
          int ind = n+1;
          for(int i = 0; i <= n; i++)
          {
              if(map.containsKey(i))
              {
                  ans[i] = map.get(i);
              }
              if(i > 0)
              {
                 // System.out.println(Arrays.toString(dp));
                  if(!map.containsKey(i-1))
                  {
                      int p = Arrays.binarySearch(arr, i-1);
                      p = -p-1;
                      //pl(i+" "+p+" "+(i-1));
                      if(p <= i-1)
                      {
                          ans[i] = -1;
                          ind = i;
                          break;
                      }
                      else
                      {
                          if(stk.isEmpty())
                          {
                              ans[i] = -1;
                              ind = i;
                              break;
                          }
                          else{
                              int top = stk.pop();
                              dp[i] += (i-1) - top;
                          }
                      }
                  }
                  dp[i] += dp[i-1];
              }
              ans[i] += dp[i];
              while(map.containsKey(i) && map.get(i) > 1)
              {
                  stk.add(i);
                  map.put(i, map.get(i)-1);
              }
              
          }
  //        pl(ind);
          for(int i = ind; i <= n; i++)
          {
              ans[i] = -1;
          }
          for(long ele : ans)
          {
              System.out.print(ele+" ");
          }
          pl();
      }
      /////////////////////////////////////////////////////////////////////////////////
  
      static class FenwickTree { // Binary Index Tree
  
          int[] tree;
          static int size;
  
          public FenwickTree(int size) {
              this.size = size;
              tree = new int[size + 5];
          }
  
          public void add(int i, int val) {
              while (i <= size) {
                  tree[i] += val;
                  i += i & -i;  // adding the decimal value of the last set bit.
              }
          }
  
          public int sum(int i) {
              int res = 0;
              while (i >= 1) {
                  res += tree[i];
                  i -= i & -i; // deleting the last set bit
              }
              return res;
          }
  
          public int sum(int l, int r) {
              return sum(r) - sum(l - 1);
          }
      }
  
      /////////////////////////////////////////////////////////////////////////////////
      static int[] nextIntArray(int n) {
          int[] arr = new int[n];
          int i = 0;
          while (i < n) {
              arr[i++] = ni();
          }
          return arr;
      }
  
      static long[] nextLongArray(int n) {
          long[] arr = new long[n];
          int i = 0;
          while (i < n) {
              arr[i++] = nl();
          }
          return arr;
      }
  
      static int[] nextIntArray1(int n) {
          int[] arr = new int[n + 1];
          int i = 1;
          while (i <= n) {
              arr[i++] = ni();
          }
          return arr;
      }
  
      /////////////////////////////////////////////////////////////////////////////////
      static int ni() {
          return sc.nextInt();
      }
  
      static long nl() {
          return sc.nextLong();
      }
  
      static double nd() {
          return sc.nextDouble();
      }
  
      /////////////////////////////////////////////////////////////////////////////////
      static void pl() {
          pw.println();
      }
  
      static void p(Object o) {
          pw.print(o + " ");
      }
  
      static void pl(Object o) {
          pw.println(o);
      }
  
  /////////////////////////////////////////////////////////////////////////////////
      static void print(Object s) {
          System.out.println(s);
      }
  
      /////////////////////////////////////////////////////////////////////////////////
      //Arrays.sort(arr(type int)) uses quick sort whose worst case time complexity is O(N^2) in case of sorted array.
      // we can either use Object (Integer, Long)[Here it will use merge sort] array or suffle the array before sorting
      // or we can use Collections.sort(list))
      public static long[] sort(long arr[]){
              List<Long> list = new ArrayList<>();
              for(long n : arr){list.add(n);}
              Collections.sort(list);
              for(int i=0;i<arr.length;i++){
                  arr[i] = list.get(i);
              }
              return arr;
          }
   
          public static int[] sort(int arr[]){
              List<Integer> list = new ArrayList<>();
              for(int n : arr){list.add(n);}
              Collections.sort(list);
              for(int i=0;i<arr.length;i++){
                  arr[i] = list.get(i);
              }
              return arr;
          }
      /////////////////////////////////////////////////////////////////////////////////
      //-----------HritikScanner class for faster input----------//
      static class HritikScanner {
  
          BufferedReader br;
          StringTokenizer st;
  
          public HritikScanner() {
              br = new BufferedReader(new InputStreamReader(System.in));
          }
  
          String next() {
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
  
          double nextDouble() {
              return Double.parseDouble(next());
          }
  
          String nextLine() {
              String str = "";
              try {
                  str = br.readLine();
              } catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
      }
  
      //////////////////////////////////////////////////////////////////
      public static class Pair implements Comparable<Pair> {
  
          int num, row, col;
  
          Pair(int num, int row, int col) {
              this.num = num;
              this.row = row;
              this.col = col;
          }
  
          public int get1() {
              return num;
          }
  
          public int get2() {
              return row;
          }
  
          public int compareTo(Pair A) {
              return this.num - A.num;
          }
  
          public String toString() {
              return num + " " + row + " " + col;
          }
      }
  
      //////////////////////////////////////////////////////////////////
      // Function to return gcd of a and b time complexity O(log(a+b))
      static long gcd(long a, long b) {
          if (a == 0) {
              return b;
          }
          return gcd(b % a, a);
      }
  
      // method to return LCM of two numbers 
      static long lcm(long a, long b) {
          return (a / gcd(a, b)) * b;
      }
  
      //////////////////////////////////////////////////////////////////
      static boolean isPrime(long n) {
          // Corner cases 
          if (n <= 1) {
              return false;
          }
          if (n <= 3) {
              return true;
          }
  
          // This is checked so that we can skip  
          // middle five numbers in below loop 
          if (n % 2 == 0 || n % 3 == 0) {
              return false;
          }
  
          for (long i = 5; i * i <= n; i = i + 6) {
              if (n % i == 0 || n % (i + 2) == 0) {
                  return false;
              }
          }
  
          return true;
      }
      //////////////////////////////////////////////////////////////////
  
      static boolean isPowerOfTwo(long n) {
          if (n == 0) {
              return false;
          }
  
          return (long) (Math.ceil((Math.log(n) / Math.log(2))))
                  == (long) (Math.floor(((Math.log(n) / Math.log(2)))));
      }
  
      public static long getFact(int n) {
          long ans = 1;
          while (n > 0) {
              ans *= n;
              ans %= MOD;
              n--;
          }
          return ans;
      }
  
      public static long pow(long n, int pow) {
          if (pow == 0) {
              return 1;
          }
          long temp = pow(n, pow / 2) % MOD;
          temp *= temp;
          temp %= MOD;
          if (pow % 2 == 1) {
              temp *= n;
          }
          temp %= MOD;
          return temp;
  
      }
  
      public static long nCr(int n, int r) {
          long ans = 1;
          int temp = n - r;
          while (n > temp) {
              ans *= n;
              ans %= MOD;
              n--;
          }
          ans *= pow(getFact(r) % MOD, MOD - 2) % MOD;
          ans %= MOD;
          return ans;
      }
  
      //////////////////////////////////////////////////////////////////
      // method returns Nth power of A 
      static double nthRoot(int A, int N) {
  
          // intially guessing a random number between 
          // 0 and 9 
          double xPre = Math.random() % 10;
  
          // smaller eps, denotes more accuracy 
          double eps = 0.001;
  
          // initializing difference between two 
          // roots by INT_MAX 
          double delX = 2147483647;
  
          // xK denotes current value of x 
          double xK = 0.0;
  
          // loop untill we reach desired accuracy 
          while (delX > eps) {
              // calculating current value from previous 
              // value by newton's method 
              xK = ((N - 1.0) * xPre
                      + (double) A / Math.pow(xPre, N - 1)) / (double) N;
              delX = Math.abs(xK - xPre);
              xPre = xK;
          }
  
          return xK;
      }
  
  }
  