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;
unsigned long long mod = 1e9 + 7;
unsigned long long ncr(long long int n, long long int k) {
  unsigned long long res = 1;
  if (k > n - k) k = n - k;
  for (int i = 0; i < k; ++i) {
    res *= (n - i);
    res /= (i + 1);
    res = res % mod;
  }
  return res % mod;
}
bool prime(unsigned long long num) {
  bool flag = true;
  for (unsigned long long i = 2; i * i <= num; i++) {
    if (num % i == 0) {
      flag = false;
      break;
    }
  }
  return flag;
}
long long ipow(long long base, long long exp) {
  long long result = 1;
  for (;;) {
    if (exp & 1) {
      result *= base;
      result = result % mod;
    }
    exp >>= 1;
    if (!exp) break;
    base *= base;
    base = base % mod;
  }
  return result % mod;
}
long long abs(long long a, long long b) {
  if ((a - b) > 0)
    return (a - b);
  else
    return (b - a);
}
int arr[100001];
int sz[100001];
void init(int n) {
  for (int i = 1; i <= n; i++) {
    arr[i] = i;
    sz[i] = 1;
  }
}
int _find(int x) {
  while (arr[x] != x) {
    arr[x] = arr[arr[x]];
    x = arr[x];
  }
  return x;
}
void _union(int a, int b) {
  if (sz[a] >= sz[b]) {
    arr[b] = arr[a];
    sz[a] += sz[b];
  } else {
    arr[a] = arr[b];
    sz[b] += sz[a];
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int q;
  cin >> q;
  while (q--) {
    int n;
    cin >> n;
    unordered_map<int, int> mp;
    mp.clear();
    vector<pair<int, int> > v;
    v.clear();
    for (int i = 0; i < n; i++) {
      int a;
      cin >> a;
      mp[a]++;
    }
    for (auto &x : mp) {
      v.push_back({x.second, x.first});
    }
    sort(v.begin(), v.end());
    int ans = 0;
    int st = 1000001;
    for (int i = v.size() - 1; i >= 0; i--) {
      int nst = v[i].first;
      if (nst < st) {
        ans += nst;
        st = nst;
      } else if (nst >= st && st >= 2) {
        nst = st - 1;
        ans += nst;
        st = nst;
      }
    }
    cout << ans << endl;
  }
}