#include <bits/stdc++.h>
using namespace std;
const long double pi = 3.14159265358979;
const long long int mod = 1e9 + 7;
long long int takemod(long long int a, long long int mod) {
  return ((a % mod) + mod) % mod;
}
long long int bs(long long int a[], long long int l, long long int r,
                 long long int x) {
  if (r >= l) {
    long long int mid = (l + r) / 2;
    if (a[mid] == x)
      return mid;
    else if (a[mid] > x)
      return bs(a, l, mid - 1, x);
    else
      return bs(a, mid + 1, r, x);
  }
  return -1;
}
long long int s_t_i(string second) {
  stringstream geek(second);
  long long int m;
  geek >> m;
  return m;
}
long long int power(long long int b, long long int p) {
  if (p == 0)
    return 1;
  else if (p % 2 == 0)
    return (power(b, p / 2) * (power(b, p / 2)));
  else
    return ((b) * (power(b, p - 1)));
}
bool isprime(long long int n) {
  if (n == 1) return false;
  if (n == 2 or n == 3) return true;
  long long int lim = sqrt(n);
  long long int i = 2;
  while (i <= lim) {
    if (n % i == 0) return false;
    i++;
  }
  return true;
}
vector<long long int> factors(long long int n) {
  vector<long long int> a;
  a.push_back(1);
  long long int lim = sqrt(n), i = 2;
  while (i <= lim) {
    if (n % i == 0) {
      a.push_back(i);
      if (i != n / i) a.push_back(n / i);
    }
    i++;
  }
  if (n != 1) a.push_back(n);
  sort(a.begin(), a.end());
  return a;
}
vector<pair<long long int, long long int> > pf_with_count(long long int b) {
  pair<long long int, long long int> p;
  vector<pair<long long int, long long int> > pf;
  long long int count = 0;
  while (b % 2 == 0) {
    b /= 2;
    count++;
  }
  if (count > 0) {
    p.first = 2;
    p.second = count;
    pf.push_back(p);
  }
  long long int i = 3;
  while (i <= sqrt(b)) {
    count = 0;
    while (b % i == 0) {
      b /= i;
      count++;
    }
    if (count > 0) {
      p.first = i;
      p.second = count;
      pf.push_back(p);
    }
    i += 2;
  }
  if (b > 2) {
    p.first = b;
    p.second = 1;
    pf.push_back(p);
  }
  return pf;
}
long long int bs_t_i(string second) {
  long long int l = strtoull(second.c_str(), NULL, 2);
  return l;
}
string i_to_bs(long long int n) {
  string second = bitset<5>(n).to_string();
  return second;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    long long int a[n];
    long long int i = 0;
    while (i < n) {
      cin >> a[i];
      i++;
    }
    sort(a, a + n);
    long long int sum = 0;
    i = 0;
    vector<long long int> v;
    while (i < n) {
      long long int count = 1;
      while (i < n - 1 and a[i] == a[i + 1]) {
        i++;
        count++;
      }
      v.push_back(count);
      i++;
    }
    sort(v.begin(), v.end(), greater<long long int>());
    i = 1;
    set<long long int> st;
    sum = v[0];
    st.insert(v[0]);
    while (i < v.size()) {
      if (v[i] < *st.begin()) {
        sum += v[i];
        st.insert(v[i]);
      } else {
        sum += max(*st.begin() - 1, (long long int)0);
        st.insert(max(*st.begin() - 1, (long long int)0));
      }
      i++;
    }
    cout << sum << endl;
  }
}

### Prompt

Create a solution in CPP for the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long double pi = 3.14159265358979;
const long long int mod = 1e9 + 7;
long long int takemod(long long int a, long long int mod) {
  return ((a % mod) + mod) % mod;
}
long long int bs(long long int a[], long long int l, long long int r,
                 long long int x) {
  if (r >= l) {
    long long int mid = (l + r) / 2;
    if (a[mid] == x)
      return mid;
    else if (a[mid] > x)
      return bs(a, l, mid - 1, x);
    else
      return bs(a, mid + 1, r, x);
  }
  return -1;
}
long long int s_t_i(string second) {
  stringstream geek(second);
  long long int m;
  geek >> m;
  return m;
}
long long int power(long long int b, long long int p) {
  if (p == 0)
    return 1;
  else if (p % 2 == 0)
    return (power(b, p / 2) * (power(b, p / 2)));
  else
    return ((b) * (power(b, p - 1)));
}
bool isprime(long long int n) {
  if (n == 1) return false;
  if (n == 2 or n == 3) return true;
  long long int lim = sqrt(n);
  long long int i = 2;
  while (i <= lim) {
    if (n % i == 0) return false;
    i++;
  }
  return true;
}
vector<long long int> factors(long long int n) {
  vector<long long int> a;
  a.push_back(1);
  long long int lim = sqrt(n), i = 2;
  while (i <= lim) {
    if (n % i == 0) {
      a.push_back(i);
      if (i != n / i) a.push_back(n / i);
    }
    i++;
  }
  if (n != 1) a.push_back(n);
  sort(a.begin(), a.end());
  return a;
}
vector<pair<long long int, long long int> > pf_with_count(long long int b) {
  pair<long long int, long long int> p;
  vector<pair<long long int, long long int> > pf;
  long long int count = 0;
  while (b % 2 == 0) {
    b /= 2;
    count++;
  }
  if (count > 0) {
    p.first = 2;
    p.second = count;
    pf.push_back(p);
  }
  long long int i = 3;
  while (i <= sqrt(b)) {
    count = 0;
    while (b % i == 0) {
      b /= i;
      count++;
    }
    if (count > 0) {
      p.first = i;
      p.second = count;
      pf.push_back(p);
    }
    i += 2;
  }
  if (b > 2) {
    p.first = b;
    p.second = 1;
    pf.push_back(p);
  }
  return pf;
}
long long int bs_t_i(string second) {
  long long int l = strtoull(second.c_str(), NULL, 2);
  return l;
}
string i_to_bs(long long int n) {
  string second = bitset<5>(n).to_string();
  return second;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    long long int a[n];
    long long int i = 0;
    while (i < n) {
      cin >> a[i];
      i++;
    }
    sort(a, a + n);
    long long int sum = 0;
    i = 0;
    vector<long long int> v;
    while (i < n) {
      long long int count = 1;
      while (i < n - 1 and a[i] == a[i + 1]) {
        i++;
        count++;
      }
      v.push_back(count);
      i++;
    }
    sort(v.begin(), v.end(), greater<long long int>());
    i = 1;
    set<long long int> st;
    sum = v[0];
    st.insert(v[0]);
    while (i < v.size()) {
      if (v[i] < *st.begin()) {
        sum += v[i];
        st.insert(v[i]);
      } else {
        sum += max(*st.begin() - 1, (long long int)0);
        st.insert(max(*st.begin() - 1, (long long int)0));
      }
      i++;
    }
    cout << sum << endl;
  }
}

```