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 long long mod = 1e9 + 7;
const long double eps = 1e-9;
const long long maxn = 1e5 + 1;
const long long inf = 5e18;
const long long minf = -inf;
bool solve() {
  long long n, ans_cnt = 0, ans_use = 0;
  cin >> n;
  vector<long long> cnt(n + 1, 0), use(n + 1, 0);
  for (long long i = 0, t, f; i < n; ++i) {
    cin >> t >> f;
    cnt[t]++;
    if (f) use[t]++;
  }
  priority_queue<pair<long long, long long>> have;
  multiset<long long> high;
  for (long long i = 1; i <= n; ++i) {
    if (!cnt[i])
      continue;
    else {
      have.push({cnt[i], use[i]});
    }
  }
  for (long long i = n; i > 0; --i) {
    while (!have.empty() && (have.top()).first >= i) {
      high.insert((have.top()).second);
      have.pop();
    }
    if (!high.empty()) {
      auto x = high.end();
      x = prev(x);
      ans_cnt += i;
      ans_use += min(i, *x);
      high.erase(x);
    }
  }
  cout << ans_cnt << " " << ans_use << "\n";
  return true;
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  cin >> t;
  while (t--) {
    if (solve()) {
    } else {
    }
  }
  return 0;
}