#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
const int N = 1000005;
const double PI = 4 * atan(1);
const double eps = 1e-7;
long long k, n, a, b;
long long x, y;
long long q;
map<long long, long long> maa;
map<long long, long long> man;
bool cmp(long long x, long long y) {
  if (maa[x] != maa[y])
    return maa[x] < maa[y];
  else
    return man[x] < man[y];
}
int main() {
  ios::sync_with_stdio(0);
  cin >> q;
  while (q--) {
    cin >> n;
    maa.clear();
    man.clear();
    for (int i = 0; i < n; i++) {
      cin >> x >> y;
      if (y == 1) man[x]++;
      maa[x]++;
    }
    vector<long long> v;
    for (auto u : maa) v.push_back(u.first);
    sort((v).begin(), (v).end(), cmp);
    reverse((v).begin(), (v).end());
    long long x = 0;
    long long ans = 0;
    x = maa[v[0]] + 1;
    long long ans1 = 0;
    priority_queue<pair<long long, long long> > pq;
    for (auto u : v) {
      pq.push(make_pair(maa[u], man[u]));
    }
    for (;;) {
      if (pq.empty()) break;
      pair<long long, long long> paa = pq.top();
      pq.pop();
      if (paa.first >= x) {
        pq.push(make_pair(x - 1, paa.second));
        continue;
      }
      long long u = paa.first;
      if (x == 1) break;
      if (u >= x - 1) {
        u = min(u, x - 1);
        ans1 += min(u, paa.second);
        ans += u;
        x = u;
      } else {
        x = u + 1;
        ans1 += min(u, paa.second);
        ans += u;
        x = u;
      }
    }
    cout << ans << " " << ans1 << endl;
  }
  return 0;
}

### Prompt

Please formulate a CPP solution to 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 int MOD = 1000000007;
const int N = 1000005;
const double PI = 4 * atan(1);
const double eps = 1e-7;
long long k, n, a, b;
long long x, y;
long long q;
map<long long, long long> maa;
map<long long, long long> man;
bool cmp(long long x, long long y) {
  if (maa[x] != maa[y])
    return maa[x] < maa[y];
  else
    return man[x] < man[y];
}
int main() {
  ios::sync_with_stdio(0);
  cin >> q;
  while (q--) {
    cin >> n;
    maa.clear();
    man.clear();
    for (int i = 0; i < n; i++) {
      cin >> x >> y;
      if (y == 1) man[x]++;
      maa[x]++;
    }
    vector<long long> v;
    for (auto u : maa) v.push_back(u.first);
    sort((v).begin(), (v).end(), cmp);
    reverse((v).begin(), (v).end());
    long long x = 0;
    long long ans = 0;
    x = maa[v[0]] + 1;
    long long ans1 = 0;
    priority_queue<pair<long long, long long> > pq;
    for (auto u : v) {
      pq.push(make_pair(maa[u], man[u]));
    }
    for (;;) {
      if (pq.empty()) break;
      pair<long long, long long> paa = pq.top();
      pq.pop();
      if (paa.first >= x) {
        pq.push(make_pair(x - 1, paa.second));
        continue;
      }
      long long u = paa.first;
      if (x == 1) break;
      if (u >= x - 1) {
        u = min(u, x - 1);
        ans1 += min(u, paa.second);
        ans += u;
        x = u;
      } else {
        x = u + 1;
        ans1 += min(u, paa.second);
        ans += u;
        x = u;
      }
    }
    cout << ans << " " << ans1 << endl;
  }
  return 0;
}

```