#include <bits/stdc++.h>
using namespace std;
int read() {
  char ch = 0;
  int w = 0, ff = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') ff = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    w = w * 10 + ch - '0';
    ch = getchar();
  }
  return ff * w;
}
int val[200005], val2[200005];
int sta[200005], T, x, f, n;
bool cmp(int a, int b) { return val[a] + val2[a] > val[b] + val2[b]; }
signed main() {
  T = read();
  while (T--) {
    n = read();
    for (int i = 1; i <= n; ++i) {
      x = read(), f = read();
      sta[i] = i;
      f ? val[x]++ : val2[x]++;
    }
    sort(sta + 1, sta + n + 1, cmp);
    priority_queue<int, vector<int>, less<int> > que;
    int now = 0, ans = 0, sum = 0;
    for (int i = n; i >= 1; --i) {
      while (val[sta[now + 1]] + val2[sta[now + 1]] >= i)
        now++, que.push(val[sta[now]]);
      if (que.empty()) continue;
      int p = que.top();
      que.pop();
      sum += i, ans += (i < p ? i : p);
    }
    printf("%d %d\n", sum, ans);
    for (int i = 1; i <= n; ++i) val[i] = val2[i] = 0;
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
int read() {
  char ch = 0;
  int w = 0, ff = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') ff = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    w = w * 10 + ch - '0';
    ch = getchar();
  }
  return ff * w;
}
int val[200005], val2[200005];
int sta[200005], T, x, f, n;
bool cmp(int a, int b) { return val[a] + val2[a] > val[b] + val2[b]; }
signed main() {
  T = read();
  while (T--) {
    n = read();
    for (int i = 1; i <= n; ++i) {
      x = read(), f = read();
      sta[i] = i;
      f ? val[x]++ : val2[x]++;
    }
    sort(sta + 1, sta + n + 1, cmp);
    priority_queue<int, vector<int>, less<int> > que;
    int now = 0, ans = 0, sum = 0;
    for (int i = n; i >= 1; --i) {
      while (val[sta[now + 1]] + val2[sta[now + 1]] >= i)
        now++, que.push(val[sta[now]]);
      if (que.empty()) continue;
      int p = que.top();
      que.pop();
      sum += i, ans += (i < p ? i : p);
    }
    printf("%d %d\n", sum, ans);
    for (int i = 1; i <= n; ++i) val[i] = val2[i] = 0;
  }
  return 0;
}

```