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;
long long int mod = 1000000007;
vector<long long int> initializeDiffArray(vector<long long int>& A);
void update(vector<long long int>& D, long long int l, long long int r,
            long long int x);
void getArray(vector<long long int>& A, vector<long long int>& D);
long long int min(long long int a, long long int b);
long long int max(long long int a, long long int b);
long long int gcd(long long int a, long long int b);
void swap(long long int* a, long long int* b);
long long int lcm(long long int a, long long int b);
long long int modpower(long long int x, long long int y, long long int p);
long long int power(long long int x, long long int y);
long long int modulo(long long int value, long long int m);
long long int myXOR(long long int x, long long int y);
long long int diff(long long int a, long long int b);
int main() {
  {
    ios ::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
  };
  long long int rep = 1;
  cin >> rep;
  while (rep--) {
    long long int n;
    cin >> n;
    map<long long int, long long int> m;
    for (long long int i = 1; i <= n; ++i) {
      long long int b;
      cin >> b;
      m[b]++;
    }
    long long int ans = 0;
    vector<long long int> vec;
    for (auto i : m) vec.push_back(i.second);
    sort(vec.begin(), vec.end());
    reverse(vec.begin(), vec.end());
    long long int last = vec[0];
    ans = vec[0];
    for (long long int i = 1; i <= vec.size() - 1; ++i) {
      if (last > vec[i]) {
        ans += vec[i];
        last = vec[i];
      } else {
        if (last == vec[i]) {
          last--;
          ans += last;
        } else {
          last--;
          ans += max(0, last);
        }
      }
    }
    cout << ans << '\n';
  }
  return 0;
}
long long int myXOR(long long int x, long long int y) {
  return (x | y) & (~x | ~y);
}
long long int min(long long int a, long long int b) {
  if (a > b) return b;
  return a;
}
long long int max(long long int a, long long int b) {
  if (a < b) return b;
  return a;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int diff(long long int a, long long int b) {
  if (a > b) return a - b;
  return b - a;
}
void swap(long long int* a, long long int* b) {
  long long int t = *a;
  *a = *b;
  *b = t;
}
long long int lcm(long long int a, long long int b) {
  if (a == 0) return b;
  if (b == 0) return a;
  return (a * b) / gcd(a, b);
}
long long int modpower(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int power(long long int x, long long int y) {
  long long int temp;
  if (y == 0) return 1;
  temp = power(x, y / 2);
  if (y % 2 == 0)
    return temp * temp;
  else
    return x * temp * temp;
}
long long int modulo(long long int value, long long int m) {
  long long int mod = value % m;
  if (value < 0) {
    mod += m;
  }
  return mod;
}
vector<long long int> initializeDiffArray(vector<long long int>& A) {
  long long int n = A.size();
  vector<long long int> D(n + 1);
  D[0] = A[0], D[n] = 0;
  for (long long int i = 1; i < n; i++) D[i] = A[i] - A[i - 1];
  return D;
}
void update(vector<long long int>& D, long long int l, long long int r,
            long long int x) {
  D[l] += x;
  D[r + 1] -= x;
}
void getArray(vector<long long int>& A, vector<long long int>& D) {
  for (long long int i = 0; i < A.size(); i++) {
    if (i == 0)
      A[i] = D[i];
    else
      A[i] = D[i] + A[i - 1];
  }
}