#include <bits/stdc++.h>
using namespace std;
int n, t[200010];
struct node {
  int x, f;
} a[200010];
bool cmp(node x, node y) { return x.x < y.x; }
priority_queue<int> q;
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    while (!q.empty()) q.pop();
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      int x, y;
      scanf("%d %d", &x, &y);
      t[x]++;
      a[x].x++;
      if (y) a[x].f++;
    }
    sort(t + 1, t + 1 + n);
    long long ans = t[n];
    for (int i = n - 1; i >= 1; i--) {
      if (t[i] >= t[i + 1]) {
        t[i] = t[i + 1] - 1;
      }
      if (t[i] > 0) ans += t[i];
      if (t[i] == 0) break;
    }
    printf("%lld ", ans);
    ans = 0;
    int H = n, l = n;
    sort(a + 1, a + 1 + n, cmp);
    while (1) {
      while (a[l].x == H && l) q.push(a[l].f), l--;
      if (!q.empty()) {
        ans += min(q.top(), H);
        q.pop();
      }
      H--;
      if (H == 0) break;
    }
    for (int i = 1; i <= n; i++) a[i].x = a[i].f = 0, t[i] = 0;
    printf("%lld\n", ans);
  }
}

### 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>
using namespace std;
int n, t[200010];
struct node {
  int x, f;
} a[200010];
bool cmp(node x, node y) { return x.x < y.x; }
priority_queue<int> q;
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    while (!q.empty()) q.pop();
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      int x, y;
      scanf("%d %d", &x, &y);
      t[x]++;
      a[x].x++;
      if (y) a[x].f++;
    }
    sort(t + 1, t + 1 + n);
    long long ans = t[n];
    for (int i = n - 1; i >= 1; i--) {
      if (t[i] >= t[i + 1]) {
        t[i] = t[i + 1] - 1;
      }
      if (t[i] > 0) ans += t[i];
      if (t[i] == 0) break;
    }
    printf("%lld ", ans);
    ans = 0;
    int H = n, l = n;
    sort(a + 1, a + 1 + n, cmp);
    while (1) {
      while (a[l].x == H && l) q.push(a[l].f), l--;
      if (!q.empty()) {
        ans += min(q.top(), H);
        q.pop();
      }
      H--;
      if (H == 0) break;
    }
    for (int i = 1; i <= n; i++) a[i].x = a[i].f = 0, t[i] = 0;
    printf("%lld\n", ans);
  }
}

```