#include <bits/stdc++.h>
using namespace std;
const int len = 2e5 + 10;
const double pi = acos(-1.0);
const long long mod = 998244353;
const int inf = 0x3f3f3f3f;
const int maxn = 1e9 + 5;
struct node {
  int x, f;
} s[len];
int res[len];
priority_queue<int> p;
bool cmp(node x, node y) { return x.x > y.x; }
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) s[i].x = 0, s[i].f = 0;
    for (int i = 1; i <= n; i++) {
      int x, f;
      scanf("%d%d", &x, &f);
      s[x].x++;
      if (f == 1) s[x].f++;
    }
    sort(s + 1, s + n + 1, cmp);
    while (!p.empty()) {
      p.pop();
    }
    long long ans = 0;
    res[0] = s[1].x + 1;
    for (int i = 1; i <= n; i++) {
      res[i] = min(res[i - 1] - 1, s[i].x);
      ans += res[i];
      if (res[i] == 0) break;
    }
    long long num = 0;
    int pos = 1;
    for (int i = 1; i <= n; i++) {
      if (res[i] == 0) break;
      for (; (pos <= n) && (s[pos].x >= res[i]); pos++) {
        p.push(s[pos].f);
      }
      num += min(p.top(), res[i]);
      p.pop();
    }
    printf("%lld %lld\n", ans, num);
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 len = 2e5 + 10;
const double pi = acos(-1.0);
const long long mod = 998244353;
const int inf = 0x3f3f3f3f;
const int maxn = 1e9 + 5;
struct node {
  int x, f;
} s[len];
int res[len];
priority_queue<int> p;
bool cmp(node x, node y) { return x.x > y.x; }
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) s[i].x = 0, s[i].f = 0;
    for (int i = 1; i <= n; i++) {
      int x, f;
      scanf("%d%d", &x, &f);
      s[x].x++;
      if (f == 1) s[x].f++;
    }
    sort(s + 1, s + n + 1, cmp);
    while (!p.empty()) {
      p.pop();
    }
    long long ans = 0;
    res[0] = s[1].x + 1;
    for (int i = 1; i <= n; i++) {
      res[i] = min(res[i - 1] - 1, s[i].x);
      ans += res[i];
      if (res[i] == 0) break;
    }
    long long num = 0;
    int pos = 1;
    for (int i = 1; i <= n; i++) {
      if (res[i] == 0) break;
      for (; (pos <= n) && (s[pos].x >= res[i]); pos++) {
        p.push(s[pos].f);
      }
      num += min(p.top(), res[i]);
      p.pop();
    }
    printf("%lld %lld\n", ans, num);
  }
  return 0;
}

```