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 pairsort(long long a[], long long b[], long long n) {
  pair<long long, long long> pairt[n];
  for (long long i = 0; i < n; i++) {
    pairt[i].first = a[i];
    pairt[i].second = b[i];
  }
  sort(pairt, pairt + n);
  for (long long i = 0; i < n; i++) {
    a[i] = pairt[i].first;
    b[i] = pairt[i].second;
  }
}
bool isPrime(long long n) {
  for (long long i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) return 0;
  }
  return 1;
}
long long tree[1400000];
long long a[1400000];
long long vis[1400000];
long long k = 0;
void buildtr(long long ss, long long se, long long ind) {
  if (ss == se) {
    tree[ind] = 0;
    return;
  }
  long long mid = (ss + se) / 2;
  buildtr(ss, mid, 2 * ind);
  buildtr(mid + 1, se, 2 * ind + 1);
  tree[ind] = 0;
}
void updatetr(long long ss, long long se, long long qs, long long qe,
              long long w, long long ind) {
  if (qe < ss || qs > se || vis[ind]) {
    return;
  }
  if (se <= qe && qs <= ss) {
    tree[ind] = w;
    vis[ind] = 1;
    return;
  }
  long long mid = (ss + se) / 2;
  updatetr(ss, mid, qs, qe, w, 2 * ind);
  updatetr(mid + 1, se, qs, qe, w, 2 * ind + 1);
}
void ansit(long long ss, long long se, long long ind, long long p) {
  if (!vis[ind])
    tree[ind] = p;
  else
    p = tree[ind];
  if (ss == se) {
    if (ss == tree[ind])
      cout << 0 << " ";
    else
      cout << tree[ind] + 1 << " ";
    return;
  }
  long long mid = (ss + se) / 2;
  ansit(ss, mid, 2 * ind, p);
  ansit(mid + 1, se, 2 * ind + 1, p);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long q;
  cin >> q;
  while (q--) {
    long long n, sum = 0;
    cin >> n;
    long long a[n], vis[2000000] = {0};
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
    }
    long long b[n], count = 1, cnt = 0;
    sort(a, a + n);
    for (long long i = 0; i < n - 1; i++) {
      if (a[i] == a[i + 1])
        count++;
      else {
        b[cnt++] = count;
        count = 1;
      }
    }
    b[cnt++] = count;
    sort(b, b + cnt);
    long long cur = b[cnt - 1];
    for (long long i = cnt - 1; i >= 0; i--) {
      if (b[i] < cur) cur = b[i];
      sum += cur;
      cur = cur - 1;
      if (cur == 0) break;
    }
    cout << sum << "\n";
  }
  return 0;
}