#include <bits/stdc++.h>
using namespace std;
vector<string> vec_splitter(string s) {
  s += ',';
  vector<string> res;
  while (!s.empty()) {
    res.push_back(s.substr(0, s.find(',')));
    s = s.substr(s.find(',') + 1);
  }
  return res;
}
void debug_out(vector<string> __attribute__((unused)) args,
               __attribute__((unused)) long long idx,
               __attribute__((unused)) long long LINE_NUM) {
  cerr << endl;
}
template <typename Head, typename... Tail>
void debug_out(vector<string> args, long long idx, long long LINE_NUM, Head H,
               Tail... T) {
  if (idx > 0)
    cerr << ", ";
  else
    cerr << "Line(" << LINE_NUM << ") ";
  stringstream ss;
  ss << H;
  cerr << args[idx] << " = " << ss.str();
  debug_out(args, idx + 1, LINE_NUM, T...);
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
long long dy8[] = {1, -1, 1, -1, 0, 0, -1, 1};
long long dx4[] = {0, 0, 1, -1};
long long dy4[] = {1, -1, 0, 0};
const long long MOD = 1000000007;
double sq(double x) { return x * x; }
template <typename T>
inline T Bigmod(T base, T power, T MOD) {
  T ret = 1;
  while (power) {
    if (power & 1) ret = (ret * base) % MOD;
    base = (base * base) % MOD;
    power >>= 1;
  }
  return ret;
}
inline void normal(long long &a) {
  a %= MOD;
  (a < 0) && (a += MOD);
}
inline long long modMul(long long a, long long b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  return (a * b) % MOD;
}
inline long long modAdd(long long a, long long b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  return (a + b) % MOD;
}
inline long long modSub(long long a, long long b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long modPow(long long b, long long p) {
  long long r = 1;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1;
  }
  return r;
}
inline long long modInverse(long long a) { return modPow(a, MOD - 2); }
inline long long modDiv(long long a, long long b) {
  return modMul(a, modInverse(b));
}
bool sortinrev(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return (a.first > b.first);
}
void solution() {
  long long n;
  cin >> n;
  vector<long long> arr(n), cnt(n + 1);
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
    cnt[arr[i]]++;
  }
  sort(arr.begin(), arr.end());
  stack<long long> st;
  vector<long long> ans(n + 1, -1);
  long long sum = 0;
  for (long long i = 0; i <= n; i++) {
    if (i > 0 && cnt[i - 1] == 0) {
      if (st.empty()) {
        break;
      }
      long long x = st.top();
      st.pop();
      sum += i - x - 1;
    }
    ans[i] = sum + cnt[i];
    while (i > 0 && cnt[i - 1] > 1) {
      cnt[i - 1]--;
      st.push(i - 1);
    }
  }
  for (auto it : ans) {
    cout << it << ' ';
  }
  cout << endl;
}
signed main() {
  ios::sync_with_stdio(false);
  long long t;
  t = 1;
  cin >> t;
  while (t--) {
    solution();
  }
  return 0;
}

### Prompt

In CPP, your task is to solve the following problem:

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. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
vector<string> vec_splitter(string s) {
  s += ',';
  vector<string> res;
  while (!s.empty()) {
    res.push_back(s.substr(0, s.find(',')));
    s = s.substr(s.find(',') + 1);
  }
  return res;
}
void debug_out(vector<string> __attribute__((unused)) args,
               __attribute__((unused)) long long idx,
               __attribute__((unused)) long long LINE_NUM) {
  cerr << endl;
}
template <typename Head, typename... Tail>
void debug_out(vector<string> args, long long idx, long long LINE_NUM, Head H,
               Tail... T) {
  if (idx > 0)
    cerr << ", ";
  else
    cerr << "Line(" << LINE_NUM << ") ";
  stringstream ss;
  ss << H;
  cerr << args[idx] << " = " << ss.str();
  debug_out(args, idx + 1, LINE_NUM, T...);
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
long long dy8[] = {1, -1, 1, -1, 0, 0, -1, 1};
long long dx4[] = {0, 0, 1, -1};
long long dy4[] = {1, -1, 0, 0};
const long long MOD = 1000000007;
double sq(double x) { return x * x; }
template <typename T>
inline T Bigmod(T base, T power, T MOD) {
  T ret = 1;
  while (power) {
    if (power & 1) ret = (ret * base) % MOD;
    base = (base * base) % MOD;
    power >>= 1;
  }
  return ret;
}
inline void normal(long long &a) {
  a %= MOD;
  (a < 0) && (a += MOD);
}
inline long long modMul(long long a, long long b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  return (a * b) % MOD;
}
inline long long modAdd(long long a, long long b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  return (a + b) % MOD;
}
inline long long modSub(long long a, long long b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long modPow(long long b, long long p) {
  long long r = 1;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1;
  }
  return r;
}
inline long long modInverse(long long a) { return modPow(a, MOD - 2); }
inline long long modDiv(long long a, long long b) {
  return modMul(a, modInverse(b));
}
bool sortinrev(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return (a.first > b.first);
}
void solution() {
  long long n;
  cin >> n;
  vector<long long> arr(n), cnt(n + 1);
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
    cnt[arr[i]]++;
  }
  sort(arr.begin(), arr.end());
  stack<long long> st;
  vector<long long> ans(n + 1, -1);
  long long sum = 0;
  for (long long i = 0; i <= n; i++) {
    if (i > 0 && cnt[i - 1] == 0) {
      if (st.empty()) {
        break;
      }
      long long x = st.top();
      st.pop();
      sum += i - x - 1;
    }
    ans[i] = sum + cnt[i];
    while (i > 0 && cnt[i - 1] > 1) {
      cnt[i - 1]--;
      st.push(i - 1);
    }
  }
  for (auto it : ans) {
    cout << it << ' ';
  }
  cout << endl;
}
signed main() {
  ios::sync_with_stdio(false);
  long long t;
  t = 1;
  cin >> t;
  while (t--) {
    solution();
  }
  return 0;
}

```