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 = 2e5 + 10;
int n, ans, tot, Q;
struct node {
  int c, d;
} a[maxn];
inline bool operator<(node a, node b) { return a.d < b.d; }
priority_queue<node> q;
int main() {
  scanf("%d", &Q);
  while (Q--) {
    scanf("%d", &n);
    int x, f;
    ans = tot = 0;
    for (int i = (1); i <= (n); ++i) a[i] = (node){0, 0};
    for (int i = (1); i <= (n); ++i)
      scanf("%d%d", &x, &f), a[x].c++, a[x].d += f;
    sort(a + 1, a + n + 1, [](node x, node y) { return x.c > y.c; });
    int j = 1;
    while (!q.empty()) q.pop();
    for (int i = (n); i >= (1); --i) {
      while (j <= n && a[j].c >= i) q.push(a[j]), ++j;
      if (!q.empty()) {
        node x = q.top();
        q.pop();
        ans += i;
        tot += min(x.d, i);
      }
    }
    printf("%d %d\n", ans, tot);
  }
  return 0;
}