#include <bits/stdc++.h>
using std::max;
using std::sort;
const int MAXN = 1E6 + 10;
int cc[MAXN], tong[MAXN];
int w[MAXN];
int doing() {
  int n, v = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    v = max(v, x);
    tong[x]++;
    cc[i] = x;
  }
  int tail = 0;
  for (int i = 1; i <= v; ++i) {
    if (tong[i] != 0) {
      w[++tail] = tong[i];
    }
  }
  for (int i = 1; i <= n; ++i) {
    int x = cc[i];
    tong[x]--;
  }
  sort(w + 1, w + tail + 1);
  int now = tail, bigged = 0;
  int ans = 0;
  for (int i = w[tail]; i >= 1; --i) {
    while (now > 0 && w[now] > i) {
      bigged++;
      now--;
    }
    if (now > 0 && w[now] == i) {
      ans += i;
      now--;
    } else if (bigged > 0) {
      ans += i;
      bigged--;
    }
  }
  printf("%d\n", ans);
  return 0;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t-- > 0) doing();
  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 std::max;
using std::sort;
const int MAXN = 1E6 + 10;
int cc[MAXN], tong[MAXN];
int w[MAXN];
int doing() {
  int n, v = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    v = max(v, x);
    tong[x]++;
    cc[i] = x;
  }
  int tail = 0;
  for (int i = 1; i <= v; ++i) {
    if (tong[i] != 0) {
      w[++tail] = tong[i];
    }
  }
  for (int i = 1; i <= n; ++i) {
    int x = cc[i];
    tong[x]--;
  }
  sort(w + 1, w + tail + 1);
  int now = tail, bigged = 0;
  int ans = 0;
  for (int i = w[tail]; i >= 1; --i) {
    while (now > 0 && w[now] > i) {
      bigged++;
      now--;
    }
    if (now > 0 && w[now] == i) {
      ans += i;
      now--;
    } else if (bigged > 0) {
      ans += i;
      bigged--;
    }
  }
  printf("%d\n", ans);
  return 0;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t-- > 0) doing();
  return 0;
}

```