#include <bits/stdc++.h>
using namespace std;
const int maxm = 2e5 + 10;
int cnt0[maxm], cnt1[maxm], a[maxm];
priority_queue<int> q;
bool cmp(int a, int b) { return cnt0[a] + cnt1[a] > cnt0[b] + cnt1[b]; }
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    while (!q.empty()) q.pop();
    int n, x, y;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      scanf("%d %d", &x, &y);
      if (y == 0)
        cnt0[x]++;
      else
        cnt1[x]++;
      a[i] = i;
    }
    sort(a + 1, a + n + 1, cmp);
    int ans = 0, num = 0, now = 1;
    for (int i = n; i >= 1; i--) {
      while (cnt0[a[now]] + cnt1[a[now]] >= i) {
        q.push(cnt1[a[now]]);
        now++;
      }
      if (q.empty()) continue;
      int dd = q.top();
      q.pop();
      ans += i;
      num += min(i, dd);
    }
    printf("%d %d\n", ans, num);
    for (int i = 1; i <= n; i++) cnt0[i] = cnt1[i] = 0;
  }
  return 0;
}

### Prompt

Your task is to create 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 maxm = 2e5 + 10;
int cnt0[maxm], cnt1[maxm], a[maxm];
priority_queue<int> q;
bool cmp(int a, int b) { return cnt0[a] + cnt1[a] > cnt0[b] + cnt1[b]; }
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    while (!q.empty()) q.pop();
    int n, x, y;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      scanf("%d %d", &x, &y);
      if (y == 0)
        cnt0[x]++;
      else
        cnt1[x]++;
      a[i] = i;
    }
    sort(a + 1, a + n + 1, cmp);
    int ans = 0, num = 0, now = 1;
    for (int i = n; i >= 1; i--) {
      while (cnt0[a[now]] + cnt1[a[now]] >= i) {
        q.push(cnt1[a[now]]);
        now++;
      }
      if (q.empty()) continue;
      int dd = q.top();
      q.pop();
      ans += i;
      num += min(i, dd);
    }
    printf("%d %d\n", ans, num);
    for (int i = 1; i <= n; i++) cnt0[i] = cnt1[i] = 0;
  }
  return 0;
}

```