Problem: This problem is actually a subproblem of problem G from the same contest.

There are n candies in a candy box. The type of the i-th candy is a_i (1 ≤ a_i ≤ n).

You have to prepare a gift using some of these candies with the following restriction: the numbers of candies of each type presented in a gift should be all distinct (i. e. for example, a gift having two candies of type 1 and two candies of type 2 is bad). 

It is possible that multiple types of candies are completely absent from the gift. It is also possible that not all candies of some types will be taken to a gift.

Your task is to find out the maximum possible size of the single gift you can prepare using the candies you have.

You have to answer q independent queries.

If you are Python programmer, consider using PyPy instead of Python when you submit your code.

Input

The first line of the input contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries. Each query is represented by two lines.

The first line of each query contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of candies.

The second line of each query contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n), where a_i is the type of the i-th candy in the box.

It is guaranteed that the sum of n over all queries does not exceed 2 ⋅ 10^5.

Output

For each query print one integer — the maximum possible size of the single gift you can compose using candies you got in this query with the restriction described in the problem statement.

Example

Input


3
8
1 4 8 4 5 6 3 8
16
2 1 3 3 4 3 4 4 1 3 2 2 2 4 1 1
9
2 2 4 4 4 7 7 7 7


Output


3
10
9

Note

In the first query, you can prepare a gift with two candies of type 8 and one candy of type 5, totalling to 3 candies.

Note that this is not the only possible solution — taking two candies of type 4 and one candy of type 6 is also valid.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 10;
bool is_Prime(long long a) {
  for (long long i = 2; i * i <= a; ++i)
    if (a % i == 0) return false;
  return true;
}
bool cmp(pair<long long, long long> a, pair<long long, long long> b) {
  return (a.second < b.second);
}
long long fact(long long n) {
  vector<long long> v;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      if (n / i == i)
        v.push_back(i);
      else {
        v.push_back(i);
        v.push_back(n / i);
      }
    }
  }
  return v.size();
}
void C(long long n, long long r) {
  vector<vector<long long> > c(n + 1, vector<long long>(r + 1, 0));
  for (long long i = 0; i < n + 1; i++) {
    for (long long j = 0; j <= i && j <= r; j++)
      if (j == 0 || j == i)
        c[i][j] = 1;
      else
        c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % 1000000007;
  }
}
bool is_pl(string s, int n) {
  for (long long i = 0; i < n / 2; i++) {
    if (s[i] != s[n - i - 1]) return false;
  }
  return true;
}
void solve() {
  int n;
  cin >> n;
  map<int, int> m;
  for (int i = 0; i < n; i++) {
    int temp;
    cin >> temp;
    m[temp]++;
  }
  vector<int> v;
  for (auto x : m) v.push_back(x.second);
  sort(v.begin(), v.end());
  reverse(v.begin(), v.end());
  int res = 0;
  int cur = 1000000007;
  for (int i = 0; i < ((int)(v).size()); i++) {
    if (cur == 0) break;
    if (v[i] < cur) {
      res += v[i];
      cur = v[i];
    } else {
      res += cur - 1;
      cur--;
    }
  }
  cout << res << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  ;
  int T = 1;
  cin >> T;
  while (T--) solve();
  return 0;
}