#include <bits/stdc++.h>
using namespace std;
inline void prnt(bool ok) { cout << (ok ? "YES" : "NO"); }
const int ppr[] = {257, 263};
const long long INF = 2e18;
const int inf = 2e9;
const int mod[] = {int(1e9 + 7), int(1e9 + 9)};
const int N = 1e6 + 123;
const long double pi = 3.141592653589793238462643;
const int dx[] = {1, 0, -1, 0};
const int dy[] = {0, 1, 0, -1};
long long q, n, a[N], first[N], c[N], t[N], ta[N];
bool u[N], w[N];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> q;
  while (q--) {
    cin >> n;
    for (long long i = 1; i <= n; i++)
      cin >> a[i] >> first[i], t[a[i]] += first[i], c[a[i]]++;
    multiset<long long> st[n + 5];
    for (long long i = 1; i <= n; i++)
      if (!u[a[i]]) st[c[a[i]]].insert(t[a[i]]), u[a[i]] = 1;
    long long ans = 0, cur = 0;
    multiset<long long> second;
    for (long long i = n; i >= 1; i--) {
      for (auto x : st[i]) second.insert(x);
      if (second.empty()) continue;
      long long x = (*second.rbegin());
      second.erase(second.find(x));
      ans += i;
      cur += min(x, i);
    }
    cout << ans << ' ' << cur << '\n';
    for (int i = 1; i <= n; i++) c[a[i]] = t[a[i]] = u[a[i]] = 0;
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
inline void prnt(bool ok) { cout << (ok ? "YES" : "NO"); }
const int ppr[] = {257, 263};
const long long INF = 2e18;
const int inf = 2e9;
const int mod[] = {int(1e9 + 7), int(1e9 + 9)};
const int N = 1e6 + 123;
const long double pi = 3.141592653589793238462643;
const int dx[] = {1, 0, -1, 0};
const int dy[] = {0, 1, 0, -1};
long long q, n, a[N], first[N], c[N], t[N], ta[N];
bool u[N], w[N];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> q;
  while (q--) {
    cin >> n;
    for (long long i = 1; i <= n; i++)
      cin >> a[i] >> first[i], t[a[i]] += first[i], c[a[i]]++;
    multiset<long long> st[n + 5];
    for (long long i = 1; i <= n; i++)
      if (!u[a[i]]) st[c[a[i]]].insert(t[a[i]]), u[a[i]] = 1;
    long long ans = 0, cur = 0;
    multiset<long long> second;
    for (long long i = n; i >= 1; i--) {
      for (auto x : st[i]) second.insert(x);
      if (second.empty()) continue;
      long long x = (*second.rbegin());
      second.erase(second.find(x));
      ans += i;
      cur += min(x, i);
    }
    cout << ans << ' ' << cur << '\n';
    for (int i = 1; i <= n; i++) c[a[i]] = t[a[i]] = u[a[i]] = 0;
  }
  return 0;
}

```