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>
const int big = 0x7fffffff;
using namespace std;
inline void read(int &x) {
  x = 0;
  char ch = getchar();
  int pd = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') pd = -pd;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  x *= pd;
}
inline void write(const int &x) {
  char ggg[10001];
  int s = 0;
  int tmp = x;
  if (tmp == 0) {
    putchar('0');
    return;
  }
  if (tmp < 0) {
    tmp = -tmp;
    putchar('-');
  }
  while (tmp > 0) {
    ggg[s++] = tmp % 10 + '0';
    tmp /= 10;
  }
  while (s > 0) {
    putchar(ggg[--s]);
  }
}
int n, mp[200010], mmp[200010];
int main() {
  int q;
  read(q);
  while (q--) {
    read(n);
    for (register int i = 1; i <= n; ++i) mp[i] = 0;
    int mx = 0;
    for (register int i = 1; i <= n; ++i) {
      int xx;
      read(xx);
      ++mp[xx];
      mx = (mx > xx ? mx : xx);
    }
    int mmx = 0;
    for (register int i = 1; i <= mx; ++i) {
      mmp[mp[i]]++;
      mmx = (mp[i] > mmx ? mp[i] : mmx);
    }
    int ans = 0;
    int sum = 0;
    for (register int i = mmx; i >= 1; --i) {
      if (mmp[i]) {
        ans += i;
        sum += mmp[i] - 1;
      } else {
        if (sum) {
          --sum;
          ans += i;
        }
      }
      mmp[i] = 0;
    }
    write(ans);
    puts("");
  }
  return 0;
}