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;
mt19937_64 rang(
    chrono::high_resolution_clock::now().time_since_epoch().count());
long long power(long long x, long long y) {
  long long v = 1;
  while (y > 0) {
    if (y & 1) v = v * x;
    y = y >> 1;
    x = x * x;
  }
  return v;
}
bool isPowerOfTwo(long long x) { return x && (!(x & (x - 1))); }
long long ceil(long long a, long long b) { return (a + b - 1) / b; }
bool areSame(vector<long long>& a, int n) {
  unordered_map<int, int> m;
  for (int i = 0; i < n; i++) m[a[i]]++;
  if (m.size() == 1)
    return true;
  else
    return false;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  srand(chrono::high_resolution_clock::now().time_since_epoch().count());
  long long test = 1;
  cin >> test;
  while (test--) {
    long long n;
    cin >> n;
    vector<int> a(n);
    map<int, int> mp;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      mp[a[i]]++;
    }
    vector<long long> vals;
    for (auto i : mp) {
      vals.push_back(i.second);
    }
    sort(vals.begin(), vals.end());
    reverse(vals.begin(), vals.end());
    long long ans = vals[0];
    long long temp = vals[0];
    for (int i = 1; i < vals.size(); i++) {
      temp = min(temp - 1, vals[i]);
      if (temp >= 0) ans += temp;
    }
    cout << ans << "\n";
  }
  return 0;
}