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;
bool isp[200005];
void sv() {
  long long i, j;
  memset(isp, true, sizeof(isp));
  isp[1] = false;
  for (i = 2; i * i < 200005; i++) {
    if (isp[i]) {
      for (j = i * i; j < 200005; j += i) isp[j] = false;
    }
  }
}
long long powr(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long q, n, i, j, a[200005];
  cin >> q;
  while (q--) {
    cin >> n;
    map<long long, long long> mp;
    for (i = 0; i < n; i++) {
      cin >> a[i];
      mp[a[i]]++;
    }
    auto it = mp.begin();
    vector<long long> vc;
    while (it != mp.end()) {
      vc.push_back(it->second);
      it++;
    }
    sort(vc.begin(), vc.end());
    reverse(vc.begin(), vc.end());
    long long ln = vc.size();
    long long ans = vc[0];
    long long aa = vc[0];
    for (i = 1; i < ln; i++) {
      long long z = 0;
      aa = min(vc[i], aa - 1);
      aa = max(z, aa);
      ans += aa;
    }
    cout << ans << "\n";
  }
  return 0;
}