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;
long long bigmod(long long b, long long p, long long md) {
  if (p == 0) return 1;
  if (p % 2 == 1) {
    return ((b % md) * bigmod(b, p - 1, md)) % md;
  } else {
    long long y = bigmod(b, p / 2, md);
    return (y * y) % md;
  }
}
int cnt[200005];
int f[200005];
vector<int> v[200005];
int main() {
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    int mx = 0;
    for (int i = 0; i < n; i++) {
      int a, b;
      cin >> a >> b;
      if (b) f[a]++;
      cnt[a]++;
      mx = max(mx, cnt[a]);
    }
    for (int i = 1; i <= n; i++) {
      v[cnt[i]].push_back(f[i]);
    }
    long long ans1 = 0;
    long long ans2 = 0;
    priority_queue<int> pq;
    for (int i = mx; i >= 1; i--) {
      for (auto c : v[i]) {
        pq.push(c);
      }
      if (pq.size()) {
        ans1 += i;
        ans2 += min(i, pq.top());
        pq.pop();
      }
    }
    cout << ans1 << " " << ans2 << endl;
    ;
    for (int i = 1; i <= mx; i++) v[i].clear();
    for (int i = 1; i <= n; i++) {
      cnt[i] = 0;
      f[i] = 0;
    }
  }
  return 0;
}