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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
const int mod = 1e6 + 7;
const int inf = 1e4;
const long long infL = 1e18;
int setbit(int n, int pos) { return n = n | (1 << pos); }
int resetbit(int n, int pos) { return n = n & ~(1 << pos); }
bool checkbit(int n, int pos) { return (bool)(n & (1 << pos)); }
template <typename T>
T mul(T x, T y) {
  x %= mod;
  y %= mod;
  return (x * y) % mod;
}
template <typename T>
T add(T x, T y) {
  x %= mod;
  y %= mod;
  return (x + y) % mod;
}
template <typename T>
T modPow(T x, T k, T mod) {
  if (k == 0) return 1;
  if (k & 1) {
    return ((x % mod) * modPow(x, k - 1, mod) % mod) % mod;
  } else {
    T ret = modPow(x, k / 2, mod);
    ret %= mod;
    return (ret * ret) % mod;
  }
}
template <typename T>
T ext_gcd(T num1, T num2, T& X, T& Y) {
  T r2, r1, q2, q1, x2, x1, y2, y1, x, y, r;
  x2 = 1;
  y2 = 0;
  x1 = 0;
  y1 = 1;
  for (r2 = num1, r1 = num2; r1 != 0;
       y2 = y1, y1 = y, x2 = x1, x1 = x, r2 = r1, r1 = r) {
    q1 = r2 / r1;
    x = x2 - q1 * x1;
    y = y2 - q1 * y1;
    r = r2 % r1;
  }
  X = x2;
  Y = y2;
  return r2;
}
template <typename T>
T modInv(T x, T m) {
  return modPow(x, m - 2, m);
}
template <typename T>
void print(const T& v) {
  cerr << v << ' ';
}
template <typename T1, typename... T2>
void print(const T1& first, const T2&... rest) {
  print(first);
  print(rest...);
}
template <typename F, typename S>
ostream& operator<<(ostream& os, const pair<F, S>& p) {
  return os << "[ " << p.first << ", " << p.second << " ]";
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "[ ";
  for (int i = 0; i < v.size(); i++) os << v[i] << ' ';
  return os << " ]";
}
const int M = (int)5e5;
const int mx = (int)700;
const int N = 1000;
pair<int, int> a[M + 2];
void solve() {
  priority_queue<int> pq;
  int n, x, y, tot = 0, id, cnt = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) a[i].first = 0, a[i].second = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &x, &y);
    a[x].first++;
    a[x].second += y;
  }
  sort(a + 1, a + n + 1);
  id = n;
  for (int curr = n; curr > 0; curr--) {
    while (id > 0 and a[id].first >= curr) pq.push(a[id--].second);
    if (pq.size() > 0) {
      tot += curr;
      cnt += min(curr, pq.top());
      pq.pop();
    }
  }
  printf("%d %d\n", tot, cnt);
}
int main() {
  int t = 1, c = 0;
  scanf("%d", &t);
  clock_t s = clock();
  while (t--) {
    solve();
  }
}