In fact, the problems E1 and E2 do not have much in common. You should probably think of them as two separate problems.

You are given an integer array a[1 … n] = [a_1, a_2, …, a_n].

Let us consider an empty [deque](https://tinyurl.com/pfeucbux) (double-ended queue). A deque is a data structure that supports adding elements to both the beginning and the end. So, if there are elements [3, 4, 4] currently in the deque, adding an element 1 to the beginning will produce the sequence [\color{red}{1}, 3, 4, 4], and adding the same element to the end will produce [3, 4, 4, \color{red}{1}].

The elements of the array are sequentially added to the initially empty deque, starting with a_1 and finishing with a_n. Before adding each element to the deque, you may choose whether to add it to the beginning or to the end.

For example, if we consider an array a = [3, 7, 5, 5], one of the possible sequences of actions looks like this:    1. |  add 3 to the beginning of the deque: |  deque has a sequence [\color{red}{3}] in it;   
---|---|---  
  2. |  add 7 to the end of the deque: |  deque has a sequence [3, \color{red}{7}] in it;   
  3. |  add 5 to the end of the deque: |  deque has a sequence [3, 7, \color{red}{5}] in it;   
  4. |  add 5 to the beginning of the deque: |  deque has a sequence [\color{red}{5}, 3, 7, 5] in it;   
  
Find the minimal possible number of inversions in the deque after the whole array is processed. 

An inversion in sequence d is a pair of indices (i, j) such that i < j and d_i > d_j. For example, the array d = [5, 3, 7, 5] has exactly two inversions — (1, 2) and (3, 4), since d_1 = 5 > 3 = d_2 and d_3 = 7 > 5 = d_4.

Input

The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases.

The next 2t lines contain descriptions of the test cases. 

The first line of each test case description contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — array size. The second line of the description contains n space-separated integers a_i (-10^9 ≤ a_i ≤ 10^9) — elements of the array.

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.

Output

Print t lines, each line containing the answer to the corresponding test case. The answer to a test case should be a single integer — the minimal possible number of inversions in the deque after executing the described algorithm.

Example

Input


6
4
3 7 5 5
3
3 2 1
3
3 1 2
4
-1 2 2 -1
4
4 5 1 3
5
1 3 1 3 2


Output


2
0
1
0
1
2

Note

One of the ways to get the sequence [5, 3, 7, 5] in the deque, containing only two inversions, from the initial array [3, 7, 5, 5] (the first sample test case) is described in the problem statement. 

Also, in this example, you could get the answer of two inversions by simply putting each element of the original array at the end of the deque. In this case, the original sequence [3, 7, 5, 5], also containing exactly two inversions, will be in the deque as-is.
/*
        "Everything in the universe is balanced. Every disappointment
                you face in life will be balanced by something good for you!
                        Keep going, never give up."

*/

import java.util.*;
import java.lang.*;
import java.io.*;

public class CodeChef {
    static class BIT {
        int n;
        int[] tree;

        BIT(int n) {
            this.n = n;
            tree = new int[n + 1];
        }

        private void update(int index, int value) {
            index++;
            while (index <= n) {
                tree[index] += value;
                index += index & -index;
            }
        }

        private int query(int l, int r) {  // range sum
            if (l > r) {
                return 0;
            }
            return presum(r) - presum(l - 1);
        }

        private int presum(int index) {
            int sum = 0;
            index++;
            while (index > 0) {
                sum += tree[index];
                index -= index & -index;
            }
            return sum;
        }
    }
    public static void main(String[] args) throws java.lang.Exception {
        out = new PrintWriter(new BufferedOutputStream(System.out));
        sc = new FastReader();

        int test = sc.nextInt();
        for (int t = 0; t < test; t++) {
            solve();
        }
        out.close();
    }

    private static void solve() {
        int n = sc.nextInt();
        int[] arr = new int[n];
        TreeSet<Integer> treeSet = new TreeSet<>();
        for (int i = 0; i < n; i++) {
            arr[i] = sc.nextInt();
            treeSet.add(arr[i]);
        }

        Map<Integer, Integer> map = new HashMap<>();
        for (int val : treeSet) {
            map.put(val, map.size()); // map <value , number of smaller elements present>
        }
        for (int i = 0; i < n; i++) {
            arr[i] = map.get(arr[i]); // number of elements smaller than arr[i]
        }

        BIT bitTree = new BIT(map.size());
        long res = 0;

        for (int val : arr) {
            int smallerElements = bitTree.query(0, val - 1);
            int biggerElements = bitTree.query(val + 1, map.size() - 1);

            res += Math.min(smallerElements, biggerElements); // minimizing total inversions

            bitTree.update(val, 1);
        }

        out.println(res);
     }

    public static FastReader sc;
    public static PrintWriter out;
    static class FastReader
    {
        BufferedReader br;
        StringTokenizer segTree;

        public FastReader()
        {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

        String next()
        {
            while (segTree == null || !segTree.hasMoreElements())
            {
                try
                {
                    segTree = new StringTokenizer(br.readLine());
                }
                catch (IOException  end)
                {
                    end.printStackTrace();
                }
            }
            return segTree.nextToken();
        }

        int nextInt()
        {
            return Integer.parseInt(next());
        }

        long nextLong()
        {
            return Long.parseLong(next());
        }

        double nextDouble()
        {
            return Double.parseDouble(next());
        }

        String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException end)
            {
                end.printStackTrace();
            }
            return str;
        }
    }
}