#include <bits/stdc++.h>
using namespace std;
using ll = long long;
bool cmp(pair<ll, ll> a, pair<ll, ll> b) {
  if (a.first != b.first)
    return a.first < b.first;
  else
    return a.second > b.second;
}
void er(multiset<pair<ll, ll>>& x) { x.erase(--x.end()); }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int q;
  cin >> q;
  while (q--) {
    int n;
    cin >> n;
    map<ll, ll> cnt, no;
    for (int i = 0; i < n; ++i) {
      ll t, f;
      cin >> t >> f;
      cnt[t]++;
      no[t] += f;
    }
    multiset<pair<ll, ll>> x;
    for (auto i : cnt) {
      x.insert({i.second, no[i.first]});
    }
    ll sum = (*x.rbegin()).first;
    ll key = (*x.rbegin()).second;
    ll lst = sum;
    er(x);
    while (x.size() && (*x.rbegin()).first >= lst) {
      ll cur = (*x.rbegin()).second;
      er(x);
      x.insert({lst - 1, cur});
    }
    while (x.size()) {
      auto cur = *x.rbegin();
      sum += cur.first;
      key += min(cur.first, cur.second);
      lst = cur.first;
      er(x);
      if (lst <= 1) {
        break;
      }
      while (x.size() && (*x.rbegin()).first >= lst) {
        ll cur = (*x.rbegin()).second;
        er(x);
        x.insert({lst - 1, cur});
      }
    }
    cout << sum << " " << key << "\n";
  }
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
bool cmp(pair<ll, ll> a, pair<ll, ll> b) {
  if (a.first != b.first)
    return a.first < b.first;
  else
    return a.second > b.second;
}
void er(multiset<pair<ll, ll>>& x) { x.erase(--x.end()); }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int q;
  cin >> q;
  while (q--) {
    int n;
    cin >> n;
    map<ll, ll> cnt, no;
    for (int i = 0; i < n; ++i) {
      ll t, f;
      cin >> t >> f;
      cnt[t]++;
      no[t] += f;
    }
    multiset<pair<ll, ll>> x;
    for (auto i : cnt) {
      x.insert({i.second, no[i.first]});
    }
    ll sum = (*x.rbegin()).first;
    ll key = (*x.rbegin()).second;
    ll lst = sum;
    er(x);
    while (x.size() && (*x.rbegin()).first >= lst) {
      ll cur = (*x.rbegin()).second;
      er(x);
      x.insert({lst - 1, cur});
    }
    while (x.size()) {
      auto cur = *x.rbegin();
      sum += cur.first;
      key += min(cur.first, cur.second);
      lst = cur.first;
      er(x);
      if (lst <= 1) {
        break;
      }
      while (x.size() && (*x.rbegin()).first >= lst) {
        ll cur = (*x.rbegin()).second;
        er(x);
        x.insert({lst - 1, cur});
      }
    }
    cout << sum << " " << key << "\n";
  }
  return 0;
}

```