#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
inline int read() {
  char c = getchar();
  int t = 0, f = 1;
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    t = (t << 3) + (t << 1) + (c ^ 48);
    c = getchar();
  }
  return t * f;
}
int q, n, a[maxn], vis[maxn];
struct node {
  int val, cnt;
} t[maxn];
bool cmp(node a, node b) {
  return a.val == b.val ? a.cnt > b.cnt : a.val < b.val;
}
bool operator<(node a, node b) { return a.cnt < b.cnt; }
priority_queue<node> qu;
int main() {
  q = read();
  while (q--) {
    n = read();
    while (!qu.empty()) qu.pop();
    for (int i = 1; i <= n; i++) t[i].val = t[i].cnt = 0, vis[i] = 0;
    for (int i = 1; i <= n; i++) {
      a[i] = read();
      int x = read();
      t[a[i]].val++;
      t[a[i]].cnt += x;
    }
    sort(t + 1, t + 1 + n, cmp);
    int mx = t[n].val;
    int ans = 0, tmp = 0, pos = n;
    for (int i = mx; i >= 1; i--) {
      while (t[pos].val >= i) {
        qu.push(t[pos]);
        pos--;
      }
      if (!qu.empty()) {
        node u = qu.top();
        qu.pop();
        ans += i;
        tmp += min(i, u.cnt);
      }
    }
    printf("%d %d\n", ans, tmp);
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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;
const int maxn = 2e5 + 5;
inline int read() {
  char c = getchar();
  int t = 0, f = 1;
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    t = (t << 3) + (t << 1) + (c ^ 48);
    c = getchar();
  }
  return t * f;
}
int q, n, a[maxn], vis[maxn];
struct node {
  int val, cnt;
} t[maxn];
bool cmp(node a, node b) {
  return a.val == b.val ? a.cnt > b.cnt : a.val < b.val;
}
bool operator<(node a, node b) { return a.cnt < b.cnt; }
priority_queue<node> qu;
int main() {
  q = read();
  while (q--) {
    n = read();
    while (!qu.empty()) qu.pop();
    for (int i = 1; i <= n; i++) t[i].val = t[i].cnt = 0, vis[i] = 0;
    for (int i = 1; i <= n; i++) {
      a[i] = read();
      int x = read();
      t[a[i]].val++;
      t[a[i]].cnt += x;
    }
    sort(t + 1, t + 1 + n, cmp);
    int mx = t[n].val;
    int ans = 0, tmp = 0, pos = n;
    for (int i = mx; i >= 1; i--) {
      while (t[pos].val >= i) {
        qu.push(t[pos]);
        pos--;
      }
      if (!qu.empty()) {
        node u = qu.top();
        qu.pop();
        ans += i;
        tmp += min(i, u.cnt);
      }
    }
    printf("%d %d\n", ans, tmp);
  }
  return 0;
}

```