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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct Node {
  int x, y;
};
const int N = 2e5 + 10;
Node a[N];
priority_queue<int> pq;
int main() {
  int T;
  scanf("%d", &T);
  int n, x, y, p, ans1, ans2;
  while (T--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) a[i].x = a[i].y = 0;
    for (int i = 1; i <= n; i++) {
      scanf("%d %d", &x, &y);
      a[x].x++;
      if (y) a[x].y++;
    }
    sort(a + 1, a + n + 1, [&](Node x, Node y) {
      if (x.x == y.x) return x.y < y.y;
      return x.x < y.x;
    });
    while (!pq.empty()) pq.pop();
    p = n;
    ans1 = ans2 = 0;
    for (int i = n; i >= 1; i--) {
      while (p && a[p].x >= i) pq.push(a[p--].y);
      if (!pq.empty()) {
        ans1 += i;
        ans2 += min(pq.top(), i);
        pq.pop();
      }
    }
    printf("%d %d\n", ans1, ans2);
  }
  return 0;
}