#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 5;
int q, n;
int a[MAXN], f[MAXN];
pair<int, int> num[MAXN];
bool cmp(pair<int, int> a, pair<int, int> b) {
  if (a.first == b.first) return a.second > b.second;
  return a.first < b.first;
}
void input() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d %d", &a[i], &f[i]);
    num[a[i]].first++;
    if (f[i] == 1) num[a[i]].second++;
  }
}
vector<int> numLis[MAXN];
void work() {
  int ans1 = 0, ans2 = 0;
  for (int i = 1; i <= n; ++i) {
    numLis[num[i].first].push_back(i);
  }
  priority_queue<int> hold;
  for (int i = n; i >= 1; --i) {
    for (int j : numLis[i]) {
      hold.push(num[j].second);
    }
    if (!hold.empty()) {
      ans1 += i;
      ans2 += min(i, hold.top());
      hold.pop();
    }
  }
  printf("%d %d\n", ans1, ans2);
}
void reset() {
  for (int i = 1; i <= n; ++i) {
    num[i] = pair<int, int>(0, 0);
    numLis[i].clear();
  }
}
int main() {
  scanf("%d", &q);
  while (q--) {
    input();
    work();
    reset();
  }
}

### Prompt

Generate 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>
using namespace std;
const int MAXN = 2e5 + 5;
int q, n;
int a[MAXN], f[MAXN];
pair<int, int> num[MAXN];
bool cmp(pair<int, int> a, pair<int, int> b) {
  if (a.first == b.first) return a.second > b.second;
  return a.first < b.first;
}
void input() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d %d", &a[i], &f[i]);
    num[a[i]].first++;
    if (f[i] == 1) num[a[i]].second++;
  }
}
vector<int> numLis[MAXN];
void work() {
  int ans1 = 0, ans2 = 0;
  for (int i = 1; i <= n; ++i) {
    numLis[num[i].first].push_back(i);
  }
  priority_queue<int> hold;
  for (int i = n; i >= 1; --i) {
    for (int j : numLis[i]) {
      hold.push(num[j].second);
    }
    if (!hold.empty()) {
      ans1 += i;
      ans2 += min(i, hold.top());
      hold.pop();
    }
  }
  printf("%d %d\n", ans1, ans2);
}
void reset() {
  for (int i = 1; i <= n; ++i) {
    num[i] = pair<int, int>(0, 0);
    numLis[i].clear();
  }
}
int main() {
  scanf("%d", &q);
  while (q--) {
    input();
    work();
    reset();
  }
}

```