1183_D. Candy Box (easy version)_17100
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.
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using llu = unsigned long long;
ll mod = 1e9 + 7;
ll mode = 998244353;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ll t, n, m, i, j, k;
  cin >> t;
  while (t--) {
    cin >> n;
    ll a[n];
    map<ll, ll> z;
    for (i = 0; i < n; i++) cin >> a[i], z[a[i]]++;
    vector<ll> v;
    for (i = 0; i < n; i++) {
      if (z[a[i]]) {
        v.push_back(z[a[i]]);
        z[a[i]] = 0;
      }
    }
    sort(v.begin(), v.end(), greater<ll>());
    ll ans = 0;
    k = INT_MAX;
    for (auto y : v) {
      k = min(y, k);
      ans += k;
      k--;
      if (k <= 0) break;
    }
    cout << ans << "\n";
  }
}
