#include <bits/stdc++.h>
using namespace std;
int mp[2][200010];
int a[200010];
set<int> s;
struct node {
  int xx;
  node(int _X) { xx = _X; }
  bool operator<(const node& w) const {
    int a = mp[0][xx] + mp[1][xx];
    int b = mp[0][w.xx] + mp[1][w.xx];
    if (a != b) return a < b;
    return mp[1][xx] < mp[1][w.xx];
  }
};
vector<int> v;
signed main() {
  ios::sync_with_stdio(0);
  int t;
  cin >> t;
  while (t--) {
    s.clear();
    int n;
    cin >> n;
    int res = 0;
    int res1 = 0;
    priority_queue<node> q;
    for (int i = 1; i <= n; i++) {
      int f;
      cin >> a[i] >> f;
      mp[f][a[i]]++;
      s.insert(a[i]);
    }
    for (auto x : s) q.push(node(x));
    int last = 0;
    while (q.size()) {
      auto tx = q.top();
      q.pop();
      int tt = tx.xx;
      int a = mp[0][tt];
      int b = mp[1][tt];
      if (a + b == 0) break;
      if (last != a + b) {
        last = a + b;
        res += a + b;
        res1 += b;
      } else {
        if (a == 0)
          mp[1][tt]--;
        else
          mp[0][tt]--;
        q.push(node(tt));
      }
    }
    for (int i = 1; i <= n; i++) mp[0][i] = mp[1][i] = 0;
    cout << res << " " << res1 << endl;
  }
  return 0;
}

### Prompt

Please formulate 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;
int mp[2][200010];
int a[200010];
set<int> s;
struct node {
  int xx;
  node(int _X) { xx = _X; }
  bool operator<(const node& w) const {
    int a = mp[0][xx] + mp[1][xx];
    int b = mp[0][w.xx] + mp[1][w.xx];
    if (a != b) return a < b;
    return mp[1][xx] < mp[1][w.xx];
  }
};
vector<int> v;
signed main() {
  ios::sync_with_stdio(0);
  int t;
  cin >> t;
  while (t--) {
    s.clear();
    int n;
    cin >> n;
    int res = 0;
    int res1 = 0;
    priority_queue<node> q;
    for (int i = 1; i <= n; i++) {
      int f;
      cin >> a[i] >> f;
      mp[f][a[i]]++;
      s.insert(a[i]);
    }
    for (auto x : s) q.push(node(x));
    int last = 0;
    while (q.size()) {
      auto tx = q.top();
      q.pop();
      int tt = tx.xx;
      int a = mp[0][tt];
      int b = mp[1][tt];
      if (a + b == 0) break;
      if (last != a + b) {
        last = a + b;
        res += a + b;
        res1 += b;
      } else {
        if (a == 0)
          mp[1][tt]--;
        else
          mp[0][tt]--;
        q.push(node(tt));
      }
    }
    for (int i = 1; i <= n; i++) mp[0][i] = mp[1][i] = 0;
    cout << res << " " << res1 << endl;
  }
  return 0;
}

```