#include <bits/stdc++.h>
const int MAXN = 2e5 + 1;
int q, n, a, f, place;
int res[MAXN], ans[2];
std::priority_queue<int> sum;
struct node {
  int num, cnt;
  node(int nn = 0, int cc = 0) {
    num = nn;
    cnt = cc;
  }
} candy[MAXN];
bool cmp(const node &x, const node &y) { return x.num > y.num; }
void ini(int key) {
  for (int i = 1; i <= key; ++i) {
    candy[i].num = 0;
    candy[i].cnt = 0;
    res[i] = 0;
  }
  ans[0] = ans[1] = 0;
  while (!sum.empty()) {
    sum.pop();
  }
  place = 1;
}
int main() {
  scanf("%d", &q);
  while (q--) {
    scanf("%d", &n);
    ini(n);
    for (int i = 1; i <= n; ++i) {
      scanf("%d%d", &a, &f);
      ++candy[a].num;
      if (f == 1) {
        ++candy[a].cnt;
      }
    }
    std::sort(candy + 1, candy + n + 1, cmp);
    res[0] = candy[1].num + 1;
    for (int i = 1; (i <= n) && (candy[i].num != 0) && (res[i - 1] != 0); ++i) {
      res[i] = std::min(res[i - 1] - 1, candy[i].num);
      ans[0] += res[i];
    }
    for (int i = 1; (i <= n) && (res[i] != 0); ++i) {
      for (; (place <= n) && (candy[place].num >= res[i]); ++place) {
        sum.push(candy[place].cnt);
      }
      ans[1] += std::min(sum.top(), res[i]);
      sum.pop();
    }
    printf("%d %d\n", ans[0], ans[1]);
  }
  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>
const int MAXN = 2e5 + 1;
int q, n, a, f, place;
int res[MAXN], ans[2];
std::priority_queue<int> sum;
struct node {
  int num, cnt;
  node(int nn = 0, int cc = 0) {
    num = nn;
    cnt = cc;
  }
} candy[MAXN];
bool cmp(const node &x, const node &y) { return x.num > y.num; }
void ini(int key) {
  for (int i = 1; i <= key; ++i) {
    candy[i].num = 0;
    candy[i].cnt = 0;
    res[i] = 0;
  }
  ans[0] = ans[1] = 0;
  while (!sum.empty()) {
    sum.pop();
  }
  place = 1;
}
int main() {
  scanf("%d", &q);
  while (q--) {
    scanf("%d", &n);
    ini(n);
    for (int i = 1; i <= n; ++i) {
      scanf("%d%d", &a, &f);
      ++candy[a].num;
      if (f == 1) {
        ++candy[a].cnt;
      }
    }
    std::sort(candy + 1, candy + n + 1, cmp);
    res[0] = candy[1].num + 1;
    for (int i = 1; (i <= n) && (candy[i].num != 0) && (res[i - 1] != 0); ++i) {
      res[i] = std::min(res[i - 1] - 1, candy[i].num);
      ans[0] += res[i];
    }
    for (int i = 1; (i <= n) && (res[i] != 0); ++i) {
      for (; (place <= n) && (candy[place].num >= res[i]); ++place) {
        sum.push(candy[place].cnt);
      }
      ans[1] += std::min(sum.top(), res[i]);
      sum.pop();
    }
    printf("%d %d\n", ans[0], ans[1]);
  }
  return 0;
}

```