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;
void c_p_c() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i += 6) {
    if (n % i == 0 || n % (i + 2) == 0) return false;
  }
  return true;
}
vector<long long> Sieve(long long n) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (long long i = 2; i * i <= n; i++) {
    if (prime[i]) {
      for (long long j = i * i; j <= n; j += i) {
        prime[j] = false;
      }
    }
  }
  vector<long long> primes;
  for (long long i = 2; i <= n; i++) {
    if (prime[i]) primes.push_back(i);
  }
  return primes;
}
map<long long, long long> primeFactors(long long n) {
  map<long long, long long> factors;
  while (n % 2 == 0) {
    n /= 2;
    factors[2]++;
  }
  for (long long i = 3; i <= sqrt(n); i += 2) {
    while (n % i == 0) {
      factors[i]++;
      n /= i;
    }
  }
  if (n > 2) factors[n]++;
  return factors;
}
int32_t main() {
  c_p_c();
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> arr(n + 1, 0);
    long long num;
    for (long long i = 0; i < n; i++) {
      cin >> num;
      arr[num]++;
    }
    sort(arr.begin(), arr.end(), greater<long long>());
    long long ans = arr[0], prev = arr[0];
    for (long long i = 1; i <= n; i++) {
      if (prev == 0) break;
      if (arr[i] >= prev) {
        ans += prev - 1;
        prev--;
      } else {
        ans += arr[i];
        prev = arr[i];
      }
    }
    cout << ans << '\n';
  }
}