#include <bits/stdc++.h>
#pragma G++ optimize(2)
using namespace std;
inline long long read() {
  char ch = ' ', last;
  long long ans = 0;
  while (ch < '0' || ch > '9') last = ch, ch = getchar();
  while (ch >= '0' && ch <= '9') ans = ans * 10 + int(ch - '0'), ch = getchar();
  if (last == '-') return -ans;
  return ans;
}
void write(long long x) {
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
int n, Q, MAX;
pair<int, int> cnt[200105];
priority_queue<int> q;
bool cmp(pair<int, int> i, pair<int, int> j) { return i > j; }
void init() {
  while (!q.empty()) q.pop();
  n = read();
  MAX = 0;
  memset(cnt, false, sizeof(pair<int, int>) * (n + 4));
  for (int i = 1; i <= n; i++) {
    int x = read(), y = read();
    cnt[x].first++, cnt[x].second += y, MAX = max(MAX, x);
  }
  sort(cnt + 1, cnt + MAX + 1, cmp);
}
int main() {
  Q = read();
  while (Q--) {
    init();
    int ans1 = 0, ans2 = 0, now = 0;
    for (int i = cnt[1].first; i >= 1; i--) {
      while (cnt[now + 1].first == i && now + 1 <= MAX)
        q.push(cnt[++now].second);
      if (!q.empty()) ans1 += i, ans2 += min(i, q.top()), q.pop();
    }
    cout << ans1 << ' ' << ans2 << endl;
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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>
#pragma G++ optimize(2)
using namespace std;
inline long long read() {
  char ch = ' ', last;
  long long ans = 0;
  while (ch < '0' || ch > '9') last = ch, ch = getchar();
  while (ch >= '0' && ch <= '9') ans = ans * 10 + int(ch - '0'), ch = getchar();
  if (last == '-') return -ans;
  return ans;
}
void write(long long x) {
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
int n, Q, MAX;
pair<int, int> cnt[200105];
priority_queue<int> q;
bool cmp(pair<int, int> i, pair<int, int> j) { return i > j; }
void init() {
  while (!q.empty()) q.pop();
  n = read();
  MAX = 0;
  memset(cnt, false, sizeof(pair<int, int>) * (n + 4));
  for (int i = 1; i <= n; i++) {
    int x = read(), y = read();
    cnt[x].first++, cnt[x].second += y, MAX = max(MAX, x);
  }
  sort(cnt + 1, cnt + MAX + 1, cmp);
}
int main() {
  Q = read();
  while (Q--) {
    init();
    int ans1 = 0, ans2 = 0, now = 0;
    for (int i = cnt[1].first; i >= 1; i--) {
      while (cnt[now + 1].first == i && now + 1 <= MAX)
        q.push(cnt[++now].second);
      if (!q.empty()) ans1 += i, ans2 += min(i, q.top()), q.pop();
    }
    cout << ans1 << ' ' << ans2 << endl;
  }
  return 0;
}

```