#include <bits/stdc++.h>
using namespace std;
template <class T>
int getbit(T s, int i) {
  return (s >> i) & 1;
}
template <class T>
T onbit(T s, int i) {
  return s | (T(1) << i);
}
template <class T>
T offbit(T s, int i) {
  return s & (~(T(1) << i));
}
template <class T>
int cntbit(T s) {
  return __builtin_popcount(s);
}
template <class T>
inline T gcd(T a, T b) {
  T r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
long long q, n, dem[200005][2];
vector<long long> v[200005];
multiset<long long> m;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> q;
  while (q--) {
    m.clear();
    cin >> n;
    for (long long i = 1, _b = n; i <= _b; i++)
      dem[i][0] = dem[i][1] = 0, v[i].clear();
    for (long long i = 1, _b = n; i <= _b; i++) {
      long long x, y;
      cin >> x >> y;
      dem[x][y]++;
    }
    for (long long i = 1, _b = n; i <= _b; i++)
      v[dem[i][0] + dem[i][1]].push_back(dem[i][1]);
    long long res1 = 0, res2 = 0;
    for (long long i = n, _b = 1; i >= _b; i--) {
      for (auto x : v[i]) m.insert(x);
      if (m.size()) {
        res1 += i;
        auto it = *m.rbegin();
        res2 += min(i, it);
        m.erase(m.find(it));
      }
    }
    cout << res1 << " " << res2 << "\n";
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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>
using namespace std;
template <class T>
int getbit(T s, int i) {
  return (s >> i) & 1;
}
template <class T>
T onbit(T s, int i) {
  return s | (T(1) << i);
}
template <class T>
T offbit(T s, int i) {
  return s & (~(T(1) << i));
}
template <class T>
int cntbit(T s) {
  return __builtin_popcount(s);
}
template <class T>
inline T gcd(T a, T b) {
  T r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
long long q, n, dem[200005][2];
vector<long long> v[200005];
multiset<long long> m;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> q;
  while (q--) {
    m.clear();
    cin >> n;
    for (long long i = 1, _b = n; i <= _b; i++)
      dem[i][0] = dem[i][1] = 0, v[i].clear();
    for (long long i = 1, _b = n; i <= _b; i++) {
      long long x, y;
      cin >> x >> y;
      dem[x][y]++;
    }
    for (long long i = 1, _b = n; i <= _b; i++)
      v[dem[i][0] + dem[i][1]].push_back(dem[i][1]);
    long long res1 = 0, res2 = 0;
    for (long long i = n, _b = 1; i >= _b; i--) {
      for (auto x : v[i]) m.insert(x);
      if (m.size()) {
        res1 += i;
        auto it = *m.rbegin();
        res2 += min(i, it);
        m.erase(m.find(it));
      }
    }
    cout << res1 << " " << res2 << "\n";
  }
  return 0;
}

```