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 INF = 1e9 + 7;
const int mod = 1e9 + 7;
long long int powmod(long long int a, long long int b) {
  long long int res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
int nxt() {
  int x;
  scanf("%d", &x);
  return x;
}
int main() {
  int tc = nxt();
  while (tc--) {
    int n = nxt();
    map<int, int> m, c;
    for (int i = 0; i < n; i++) {
      int a = nxt(), b = nxt();
      m[a]++;
      if (b) c[a]++;
    }
    int high = 0;
    vector<pair<int, int> > order;
    for (auto x : m) {
      high = max(high, x.second);
      order.push_back({x.second, c[x.first]});
    }
    sort((order).begin(), (order).end());
    reverse((order).begin(), (order).end());
    int ptr = 0;
    int a1 = 0, a2 = 0;
    multiset<int> cnt;
    for (int height = high; height > 0; height--) {
      while (ptr < order.size() && order[ptr].first >= height) {
        cnt.insert(order[ptr].second);
        ptr++;
      }
      if (cnt.size()) {
        a1 += height;
        int tmp = *cnt.rbegin();
        a2 += min(height, tmp);
        cnt.erase(cnt.find(tmp));
      }
    }
    printf("%d %d\n", a1, a2);
  }
  return 0;
}