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;
const int MAXN = 2 * 100 * 1000;
struct candy {
  int type, num, f;
  bool operator<(const candy &c) const {
    if (this->num != c.num) return this->num > c.num;
    if (this->f != c.f) return this->f < c.f;
    return this->type < c.type;
  }
  candy(int type, int num, int f) {
    this->type = type;
    this->num = num;
    this->f = f;
  }
};
int q, n, a[MAXN], b[MAXN];
set<int> ind;
void solveQuery();
int main() {
  ios::sync_with_stdio(false);
  cin >> q;
  for (int i = 0; i < q; i++) {
    while (ind.size()) {
      int tmp = *ind.begin();
      ind.erase(tmp);
      a[tmp] = b[tmp] = 0;
    }
    cin >> n;
    for (int j = 0; j < n; j++) {
      int t, f;
      cin >> t;
      a[t - 1]++;
      if (!f) b[t - 1]++;
      ind.insert(t - 1);
    }
    solveQuery();
  }
  return 0;
}
void solveQuery() {
  set<candy> s;
  for (set<int>::iterator it = ind.begin(); it != ind.end(); it++)
    s.insert(candy(*it, a[*it], b[*it]));
  int ans = 0, minF = 0, cnt = s.begin()->num;
  while (cnt && s.size()) {
    candy c = *s.begin();
    if (c.num > cnt) {
      s.erase(c);
      c.num--;
      if (c.f) c.f--;
      s.insert(c);
    } else {
      ans += c.num;
      minF += c.f;
      s.erase(c);
      cnt = c.num - 1;
      ;
    }
  }
  cout << ans << endl;
}