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 maxn = 1e6 + 7;
int q, n, a[maxn];
map<int, int> mp;
multiset<int> qq;
int fa[maxn];
void init() {
  mp.clear();
  qq.clear();
  for (int i = 0; i <= n; i++) fa[i] = i;
}
int findfa(int x) {
  if (fa[x] == x) return x;
  return fa[x] = findfa(fa[x]);
}
int main() {
  scanf("%d", &q);
  while (q--) {
    scanf("%d", &n);
    init();
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      ++mp[a[i]];
    }
    for (auto &it : mp) {
      qq.insert(it.second);
    }
    int sum = 0;
    int last = 0;
    for (auto &it : qq) {
      int now;
      if (it != last) {
        now = it;
        fa[now] = it - 1;
      } else {
        now = findfa(it);
        fa[now] = max(now - 1, 0);
      }
      sum += now;
      last = it;
    }
    printf("%d\n", sum);
  }
  return 0;
}