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 double eps = 1e-10;
const double pi = 3.14159265358979323846264338;
inline long long multi(long long x, long long y) {
  long long tmp =
      (x * y -
       (long long)((long double)x / 998244353 * y + 1.0e-8) * 998244353);
  return tmp < 0 ? tmp + 998244353 : tmp;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    n = n + 1;
    int cnt[n];
    int vis[n];
    memset(cnt, 0, sizeof(cnt));
    memset(vis, 0, sizeof(vis));
    for (int i = 1; i < n; i++) {
      int x;
      scanf("%d", &x);
      cnt[x]++;
    }
    long long ans = 0;
    sort(cnt, cnt + n, greater<int>());
    for (int i = 0; cnt[i] > 0; i++) {
      while (vis[cnt[i]]) {
        cnt[i]--;
      }
      if (cnt[i] > 0) {
        vis[cnt[i]] = 1;
        ans += cnt[i];
      }
    }
    printf("%lld\n", ans);
  }
  return 0;
}