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 check(long long n) {
  long long j = 0, p = 0;
  while (n > 0) {
    j = n % 10;
    p += j;
    n /= 10;
  }
  if (p % 4 == 0) return 1;
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long i, j, k = 0, m = 0, n, t, f = 0, x = 0, z = 0, y = 0, ans = 0,
                  mn = 10000000009, mx = 0;
  cin >> t;
  while (t--) {
    mx = 0;
    cin >> n;
    long long a[n + 9];
    map<long long, long long> mp, np;
    for (i = 0; i < n; i++) {
      cin >> a[i];
      mp[a[i]]++;
      mx = max(mx, mp[a[i]]);
    }
    sort(a, a + n);
    vector<pair<long long, long long> > v;
    for (i = 0; i < n;) {
      v.push_back({mp[a[i]], a[i]});
      i += mp[a[i]];
    }
    sort(v.begin(), v.end());
    reverse(v.begin(), v.end());
    ans = 0;
    k = mx;
    for (i = 0; i < v.size(); i++) {
      if (v[i].first >= k) {
        np[k] = 1;
        k--;
      } else {
        k = v[i].first;
        np[k] = 1;
        k--;
      }
    }
    for (i = mx; i >= 0; i--) {
      if (np[i]) {
        ans += i;
      }
    }
    cout << ans << "\n";
  }
  return 0;
}