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;
bool isPrime(int n) {
  if (n <= 1) {
    return false;
  }
  long long c[100];
  c[0] = 1;
  for (int i = 0; i < n; c[0] = -c[0], i++) {
    c[1 + i] = 1;
    for (int j = i; j > 0; j--) c[j] = c[j - 1] - c[j];
  }
  c[0]++, c[n]--;
  int i = n;
  while (i-- && c[i] % n == 0)
    ;
  return i < 0;
}
int GCD(int a, int b) {
  if (b == 0) return a;
  return GCD(b, a % b);
}
int MostCommonInt(vector<int> v) {
  int max = 0, mostvalue = v[0];
  for (int i = 0; i < v.size(); i++) {
    int co = (int)count(v.begin(), v.end(), v[i]);
    if (co > max) {
      max = co;
      mostvalue = v[i];
    }
  }
  return mostvalue;
}
void solve() {
  int n;
  cin >> n;
  map<int, int> MAP;
  for (int i = 0; i < n; i++) {
    int a = 0;
    cin >> a;
    MAP[a] += 1;
  }
  map<int, int>::iterator it = MAP.begin();
  vector<int> v;
  for (it = MAP.begin(); it != MAP.end(); it++) {
    int num = (*it).second;
    bool b = false;
    while (num > 0 && b == false) {
      if (count(v.begin(), v.end(), num) == 0) {
        v.push_back(num);
        b = true;
      } else {
        num -= 1;
      }
    }
  }
  long long sum = 0;
  for (int i = 0; i < v.size(); i++) {
    sum += v[i];
  }
  cout << sum << endl;
}
int main() {
  int T;
  cin >> T;
  for (int i = 0; i < T; i++) {
    solve();
  }
}