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 JAVA):

Solution:
"use strict";

var q = +readline()

while (q--) {
    var n = +readline();
    var A = readline().split(' ').map(value => +value);
    var M = {};

    for (var i = 0; i < A.length; ++i) {
        var value = A[i];
        M[value] = (M[value] || 0) + 1;
    }

    var R = {};
    for (var key in M) {
        if (M.hasOwnProperty(key)) {
            var value = M[key];
            R[value] = (R[value] || 0) + 1;
        }
    }
    
    var L = Object.keys(R).map(value => +value).sort(function(a, b){return b - a});
    var result = 0;
    var lastMax = L[0];
    loop:for (var i = 0; i < L.length; ++i) {
        if (lastMax > L[i]) {
            lastMax = L[i];
        }
        var count = +R[L[i]];

        while (count--) {
            result += lastMax;
            lastMax--;
            if (lastMax === 0) {
                break loop;
            }
        }
    }
    
    write(result, '\n');
}