#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 10;
const int INF = 0x3f3f3f3f;
struct node {
  int a, b;
} num[maxn];
bool cmp1(node a, node b) {
  if (b.a == a.a) return a.b < b.b;
  return a.a < b.a;
}
int n, u, ans, v, ans2;
priority_queue<int> q;
int main() {
  int qq;
  scanf("%d", &qq);
  while (qq--) {
    scanf("%d", &n);
    for (int i = 0; i <= n; i++) num[i].a = num[i].b = 0;
    while (!q.empty()) q.pop();
    ans = 0;
    ans2 = 0;
    for (int i = 1; i <= n; i++) {
      scanf("%d", &u);
      scanf("%d", &v);
      num[u].a++;
      if (v == 1) num[u].b++;
    }
    sort(num + 1, num + 1 + n, cmp1);
    int nw = n;
    int last = n + 1;
    for (int i = n; i >= 1; i--) {
      if (nw > num[i].a) {
        nw = num[i].a;
        ans += nw;
        ans2 += num[i].b;
        last = i;
      } else {
        int t = last;
        for (int j = last - 1; j; j--) {
          if (num[j].a >= nw) {
            t = j;
            q.push(num[j].b);
          } else
            break;
        }
        last = t;
        int top = q.top();
        q.pop();
        ans += nw;
        ans2 += min(top, nw);
      }
      nw--;
      if (nw <= 0) break;
    }
    printf("%d %d\n", ans, ans2);
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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 maxn = 1e6 + 10;
const int INF = 0x3f3f3f3f;
struct node {
  int a, b;
} num[maxn];
bool cmp1(node a, node b) {
  if (b.a == a.a) return a.b < b.b;
  return a.a < b.a;
}
int n, u, ans, v, ans2;
priority_queue<int> q;
int main() {
  int qq;
  scanf("%d", &qq);
  while (qq--) {
    scanf("%d", &n);
    for (int i = 0; i <= n; i++) num[i].a = num[i].b = 0;
    while (!q.empty()) q.pop();
    ans = 0;
    ans2 = 0;
    for (int i = 1; i <= n; i++) {
      scanf("%d", &u);
      scanf("%d", &v);
      num[u].a++;
      if (v == 1) num[u].b++;
    }
    sort(num + 1, num + 1 + n, cmp1);
    int nw = n;
    int last = n + 1;
    for (int i = n; i >= 1; i--) {
      if (nw > num[i].a) {
        nw = num[i].a;
        ans += nw;
        ans2 += num[i].b;
        last = i;
      } else {
        int t = last;
        for (int j = last - 1; j; j--) {
          if (num[j].a >= nw) {
            t = j;
            q.push(num[j].b);
          } else
            break;
        }
        last = t;
        int top = q.top();
        q.pop();
        ans += nw;
        ans2 += min(top, nw);
      }
      nw--;
      if (nw <= 0) break;
    }
    printf("%d %d\n", ans, ans2);
  }
  return 0;
}

```