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;
long long max(long long a, long long b) {
  if (a > b)
    return a;
  else
    return b;
}
long long min(long long a, long long b) {
  if (a < b)
    return a;
  else
    return b;
}
long long fx[] = {1, -1, 0, 0};
long long fy[] = {0, 0, 1, -1};
long long A[300005], B[300005], C[300005], D[200005], visited[300005],
    level[300005];
char CH[1000][1000], ch;
vector<long long> V;
vector<long long> V1;
vector<long long> G;
int main() {
  long long T, N, E, u, v, i, j, k, sum = 0, ck = 0, x, y;
  string S, S1;
  double d1, d2, d3;
  cin >> T;
  for (i = 1; i <= T; i++) {
    cin >> N;
    for (k = 0; k <= N; k++) A[k] = 0;
    long long ans = 0;
    for (j = 1; j <= N; j++) {
      cin >> u;
      A[u]++;
    }
    sort(A + 1, A + N + 1);
    long long mx = N + 1;
    ans = 0;
    for (j = N; j >= 0 && mx > 1; j--) {
      v = min(A[j], mx - 1);
      ans += v;
      mx = v;
    }
    cout << ans << endl;
  }
  return 0;
}