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;
const int mod = 1e+9 + 7;
int gcd(int a, int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long binomialCoeff(long long n, long long k) {
  long long res = 1;
  if (n < k) return 0;
  if (k > n - k) k = n - k;
  for (long long i = 0; i < k; ++i) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
class mysort {
 public:
  bool operator()(const pair<int, int> &a, const pair<int, int> &b) {
    if (a.second == b.second) return a.first < b.first;
    return a.second < b.second;
  }
};
void solve() {
  int n, x, w;
  long long sum1 = 0, sum2 = 0;
  cin >> n;
  pair<int, int> rec[n + 1];
  for (int i = 0; i < n + 1; i++) {
    rec[i].first = rec[i].second = 0;
  }
  for (int i = 0; i < n; i++) {
    cin >> x >> w;
    rec[x].first++;
    rec[x].second += w;
  }
  multimap<int, int> hash(rec, rec + n + 1);
  hash.erase(0);
  multiset<pair<int, int>, mysort> mset;
  vector<pair<int, int>> v;
  auto stop = hash.begin();
  auto it = hash.end();
  it--;
  int val = it->first, flag = 0, fix;
  while (true) {
    while (it != stop && it->first == val) {
      mset.insert({it->first, it->second});
      it--;
    }
    fix = it->first;
    if (it == stop && it->first == val) {
      mset.insert({it->first, it->second});
      flag = 1;
      fix--;
    }
    while (!mset.empty() && val > fix) {
      auto itr = mset.end();
      itr--;
      v.push_back({val, min(itr->second, val)});
      mset.erase(itr);
      val--;
    }
    if (flag) break;
    val = it->first;
  }
  while (val >= 1 && !mset.empty()) {
    auto itr = mset.end();
    itr--;
    v.push_back({val, min(itr->second, val)});
    mset.erase(itr);
    val--;
  }
  for (__typeof(v.begin()) it = v.begin(); it != v.end(); it++) {
    sum1 += it->first;
    sum2 += it->second;
  }
  cout << sum1 << " " << sum2 << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) solve();
}