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:
(function () {
    var read = {
        number: function () {
            return +readline();
        },
        arr: function (divider) {
            return readline().split(divider ? divider : '');
        },
        arrNumber: function (divider) {
            return readline()
                .split(divider ? divider : '')
                .map(item => +item);
        }
    };

    var q = read.number();
    var res = '';
    while (q) {
        q--;

        var n = read.number();
        var a = read.arrNumber(' ');
        var t = {};
        a.forEach(i => {
            if(t[i]) {
                t[i]++;
            }
            else {
                t[i] = 1;
            }
        });

        var unic = {};
        for(var type in t) {
            if(unic[t[type]]) {
                unic[t[type]]++;
            }
            else {
                unic[t[type]] = 1;
            }
        }

        var f = Object.keys(unic).sort((a,b) => b - a);

        var count = 0;
        var max = +f[0];

        f.forEach(key => {
            if(max === 0) {
                return;
            }
            if(max > key) {
                max = +key;
            }
            while(unic[key]) {
                if(max === 0) {
                    break;
                }
                unic[key]--;
                count += max;
                max--;
            }
        });
        res += count + '\n';
    }
    print(res);
}());