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;
inline int bit(int x, int i) { return (x >> i) & 1; }
void CountTriangles(vector<int> A) {
  int n = A.size();
  sort(A.begin(), A.end());
  int count = 0;
  for (int i = n - 1; i >= 1; i--) {
    int l = 0, r = i - 1;
    while (l < r) {
      if (A[l] + A[r] > A[i]) {
        count += r - l;
        r--;
      } else
        l++;
    }
  }
  cout << "No of possible solutions: " << count;
}
void dfs(int i, vector<vector<int>> gr, vector<bool> &vis,
         vector<int> &output) {
  vis[i] = true;
  for (int nodes : gr[i]) {
    if (!vis[nodes]) {
      dfs(nodes, gr, vis, output);
    }
  }
  output.push_back(i);
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int q;
  cin >> q;
  while (q--) {
    long long int n;
    cin >> n;
    map<int, int> v;
    for (long long int i = 0; i < n; i++) {
      int val;
      cin >> val;
      v[val]++;
    }
    vector<int> s;
    for (__typeof((v).begin()) it = (v).begin(); it != (v).end(); it++) {
      s.push_back(it->second);
    }
    sort(s.rbegin(), s.rend());
    int sum = s[0], top = s[0];
    for (long long int i = 1; i < s.size(); i++) {
      if (s[i] >= top) {
        if (top > 1) {
          top -= 1;
          sum += top;
        }
      } else {
        sum += s[i];
        top = s[i];
      }
    }
    cout << sum << endl;
  }
  return 0;
}