#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
using namespace std;
long long int MOD = 1000000007;
double eps = 1e-12;
void solve();
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> v(n);
  unordered_map<long long int, long long int> mp;
  for (long long int i = 0; i < n; i++) {
    cin >> v[i];
    mp[v[i]]++;
  }
  vector<long long int> ans;
  for (auto i : mp) ans.push_back(i.second);
  sort(ans.begin(), ans.end(), greater<long long int>());
  long long int ans1 = ans[0];
  long long int pre = ans[0];
  for (int i = 1; i < ans.size(); i++) {
    if (ans[i] >= pre)
      pre--;
    else
      pre = ans[i];
    if (pre <= 0) break;
    ans1 += pre;
  }
  cout << ans1 << endl;
}

### Prompt

Your challenge is to write a cpp solution to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
using namespace std;
long long int MOD = 1000000007;
double eps = 1e-12;
void solve();
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> v(n);
  unordered_map<long long int, long long int> mp;
  for (long long int i = 0; i < n; i++) {
    cin >> v[i];
    mp[v[i]]++;
  }
  vector<long long int> ans;
  for (auto i : mp) ans.push_back(i.second);
  sort(ans.begin(), ans.end(), greater<long long int>());
  long long int ans1 = ans[0];
  long long int pre = ans[0];
  for (int i = 1; i < ans.size(); i++) {
    if (ans[i] >= pre)
      pre--;
    else
      pre = ans[i];
    if (pre <= 0) break;
    ans1 += pre;
  }
  cout << ans1 << endl;
}

```