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 N = 200004;
int a[N], b[N], n, p = 0;
map<int, pair<int, int> > mp;
vector<int> g[N];
pair<int, int> pa[N];
pair<int, int> f() {
  int s = 0, t = 0, mx, m;
  p = 0;
  for (auto it = mp.begin(); it != mp.end(); it++) pa[p++] = it->second;
  sort(pa, pa + p, greater<pair<int, int> >());
  mx = m = pa[0].first;
  priority_queue<int> q;
  for (int i = 0; i < p; i++) {
    g[pa[i].first].push_back(pa[i].second);
  }
  for (int i = mx; i >= 1; i--) {
    for (int x : g[i]) q.push(x);
    if (!q.empty()) {
      s += i;
      t += min(q.top(), i);
      q.pop();
    }
  }
  for (int i = 1; i <= m; i++) g[i].clear();
  return pair<int, int>(s, t);
}
int main() {
  int T;
  vector<int> u, v;
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d %d", &a[i], &b[i]);
    mp.clear();
    for (int i = 1; i <= n; i++) {
      ++mp[a[i]].first;
      if (b[i]) ++mp[a[i]].second;
    }
    pair<int, int> pa = f();
    printf("%d %d\n", pa.first, pa.second);
  }
}