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;
inline int64_t max(int64_t a, int64_t b) { return (a > b) ? a : b; }
inline int64_t min(int64_t a, int64_t b) { return (a < b) ? a : b; }
inline int64_t max3(int64_t a, int64_t b, int64_t c) {
  return max(a, max(b, c));
}
inline int64_t min3(int64_t a, int64_t b, int64_t c) {
  return min(a, min(b, c));
}
int64_t gcd(int64_t a, int64_t b) { return (a % b == 0) ? b : gcd(b, a % b); }
int64_t pow3(int64_t a, int64_t b, int64_t m = 1000000007) {
  int64_t res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % m;
    a = (a * a) % m;
    b >>= 1;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int a[n];
    unordered_map<int, int> m1;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      m1[a[i]]++;
    }
    vector<pair<int, int> > p;
    for (auto i : m1) p.push_back(make_pair(i.second, i.first));
    sort(p.begin(), p.end());
    unordered_map<int, int> fre;
    for (auto i : p) fre[i.first]++;
    vector<int> out;
    for (auto i : fre)
      for (int j = 0; j < i.second; j++) out.push_back(i.first);
    sort(out.begin(), out.end());
    int x = out.size();
    for (int i = 1; i < x; i++)
      if (i < 1)
        continue;
      else if (out[i] == out[i - 1] && out[i] != 0) {
        out[i - 1]--;
        i -= 2;
      }
    int ans = 0;
    for (auto i : out) ans += i;
    cout << ans << endl;
  }
  return 0;
}