#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 50;
const int INF = 0x3f3f3f3f;
struct node {
  int num, f;
} a[maxn];
int mp[maxn], vis[maxn], v[maxn];
struct Node {
  int num, f;
  bool operator<(const Node& tmp) const {
    if (num == tmp.num) return f < tmp.f;
    return num < tmp.num;
  }
};
priority_queue<Node> q;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    int m = 0;
    for (int i = 0; i <= n; i++) {
      mp[i] = 0;
      a[i].num = 0;
      a[i].f = 0;
      vis[i] = 0;
      v[i] = 0;
    }
    for (int i = 1; i <= n; i++) {
      int k, f;
      scanf("%d%d", &k, &f);
      if (++v[k] == 1) mp[k] = ++m;
      int index = mp[k];
      a[index].num++;
      if (f == 1) a[index].f++;
    }
    long long ans = 0;
    long long ansf = 0;
    while (!q.empty()) q.pop();
    for (int i = 1; i <= m; i++) q.push({a[i].num, a[i].f});
    while (!q.empty()) {
      Node k = q.top();
      q.pop();
      if (vis[k.num] == 0) {
        vis[k.num] = 1;
        ans += k.num;
        ansf += k.f;
        continue;
      }
      while (vis[k.num] == 1 && k.num > 0) {
        k.num--;
        k.f = min(k.f, k.num);
      }
      if (k.num > 0) q.push(k);
    }
    printf("%lld %lld\n", ans, ansf);
  }
}

### 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 maxn = 2e5 + 50;
const int INF = 0x3f3f3f3f;
struct node {
  int num, f;
} a[maxn];
int mp[maxn], vis[maxn], v[maxn];
struct Node {
  int num, f;
  bool operator<(const Node& tmp) const {
    if (num == tmp.num) return f < tmp.f;
    return num < tmp.num;
  }
};
priority_queue<Node> q;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    int m = 0;
    for (int i = 0; i <= n; i++) {
      mp[i] = 0;
      a[i].num = 0;
      a[i].f = 0;
      vis[i] = 0;
      v[i] = 0;
    }
    for (int i = 1; i <= n; i++) {
      int k, f;
      scanf("%d%d", &k, &f);
      if (++v[k] == 1) mp[k] = ++m;
      int index = mp[k];
      a[index].num++;
      if (f == 1) a[index].f++;
    }
    long long ans = 0;
    long long ansf = 0;
    while (!q.empty()) q.pop();
    for (int i = 1; i <= m; i++) q.push({a[i].num, a[i].f});
    while (!q.empty()) {
      Node k = q.top();
      q.pop();
      if (vis[k.num] == 0) {
        vis[k.num] = 1;
        ans += k.num;
        ansf += k.f;
        continue;
      }
      while (vis[k.num] == 1 && k.num > 0) {
        k.num--;
        k.f = min(k.f, k.num);
      }
      if (k.num > 0) q.push(k);
    }
    printf("%lld %lld\n", ans, ansf);
  }
}

```