#include <bits/stdc++.h>
using namespace std;
long long int MOD1 = 1000000007;
long long int MOD2 = 998244353;
void yes() { cout << "YES" << endl; }
void no() { cout << "NO" << endl; }
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int lcm(long long int a, long long int b) {
  return (a * b / (gcd(a, b)));
}
void solve() {
  long long int n;
  cin >> n;
  map<long long int, long long int> make_pair;
  for (long long int i = 0; i <= n - 1; i++) {
    long long int t;
    cin >> t;
    make_pair[t]++;
  }
  map<long long int, long long int> m;
  for (auto it = make_pair.begin(); it != make_pair.end(); it++) {
    m[it->second]++;
  }
  long long int ans = 0;
  vector<bool> visited(1e6, false);
  for (auto x : m) {
    long long int num = x.first;
    long long int times = x.second;
    for (long long int i = num; i > 0; i--) {
      if (visited[i] == false && times > 0) {
        ans += i;
        times--;
        visited[i] = true;
      }
      if (times == 0) break;
    }
  }
  cout << ans << endl;
}
int main(int argc, const char* argv[]) {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int tests = 1;
  cin >> tests;
  while (tests--) {
    solve();
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long int MOD1 = 1000000007;
long long int MOD2 = 998244353;
void yes() { cout << "YES" << endl; }
void no() { cout << "NO" << endl; }
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int lcm(long long int a, long long int b) {
  return (a * b / (gcd(a, b)));
}
void solve() {
  long long int n;
  cin >> n;
  map<long long int, long long int> make_pair;
  for (long long int i = 0; i <= n - 1; i++) {
    long long int t;
    cin >> t;
    make_pair[t]++;
  }
  map<long long int, long long int> m;
  for (auto it = make_pair.begin(); it != make_pair.end(); it++) {
    m[it->second]++;
  }
  long long int ans = 0;
  vector<bool> visited(1e6, false);
  for (auto x : m) {
    long long int num = x.first;
    long long int times = x.second;
    for (long long int i = num; i > 0; i--) {
      if (visited[i] == false && times > 0) {
        ans += i;
        times--;
        visited[i] = true;
      }
      if (times == 0) break;
    }
  }
  cout << ans << endl;
}
int main(int argc, const char* argv[]) {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int tests = 1;
  cin >> tests;
  while (tests--) {
    solve();
  }
  return 0;
}

```