#include <bits/stdc++.h>
using namespace std;
int dp1[200005], dp2[200005];
priority_queue<int> pq[200005];
int main() {
  int q;
  scanf("%d", &q);
  while (q--) {
    int n, a, b;
    scanf("%d", &n);
    for (int i = 0; i <= n; i++) {
      dp1[i] = 0, dp2[i] = 0;
      while (!pq[i].empty()) pq[i].pop();
    }
    for (int i = 1; i <= n; i++) {
      scanf("%d%d", &a, &b);
      dp1[a]++;
      if (b == 1) dp2[a]++;
    }
    for (int i = 1; i <= n; i++) {
      pq[dp1[i]].push(dp2[i]);
    }
    priority_queue<int> x;
    int ans1 = 0, ans2 = 0;
    for (int i = n; i >= 1; i--) {
      while (!pq[i].empty()) {
        x.push(pq[i].top());
        pq[i].pop();
      }
      if (x.size() > 0) {
        int val = x.top();
        x.pop();
        if (val > i) val = i;
        ans1 += i;
        ans2 += val;
      }
    }
    printf("%d %d\n", ans1, ans2);
  }
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
int dp1[200005], dp2[200005];
priority_queue<int> pq[200005];
int main() {
  int q;
  scanf("%d", &q);
  while (q--) {
    int n, a, b;
    scanf("%d", &n);
    for (int i = 0; i <= n; i++) {
      dp1[i] = 0, dp2[i] = 0;
      while (!pq[i].empty()) pq[i].pop();
    }
    for (int i = 1; i <= n; i++) {
      scanf("%d%d", &a, &b);
      dp1[a]++;
      if (b == 1) dp2[a]++;
    }
    for (int i = 1; i <= n; i++) {
      pq[dp1[i]].push(dp2[i]);
    }
    priority_queue<int> x;
    int ans1 = 0, ans2 = 0;
    for (int i = n; i >= 1; i--) {
      while (!pq[i].empty()) {
        x.push(pq[i].top());
        pq[i].pop();
      }
      if (x.size() > 0) {
        int val = x.top();
        x.pop();
        if (val > i) val = i;
        ans1 += i;
        ans2 += val;
      }
    }
    printf("%d %d\n", ans1, ans2);
  }
}

```