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

public class Codechef {
	public static void main(String[] args) throws java.lang.Exception {
		FastReader in = new FastReader(System.in);
		StringBuilder sb = new StringBuilder();
		int t = 1;
		t = in.nextInt();
		while (t > 0) {
			--t;
			int n = in.nextInt();
			int arr[] = new int[n];
			TreeSet<Integer> set = new TreeSet<>();
			for(int i = 0;i<n;i++)
			{
			   arr[i] = in.nextInt();
			   set.add(arr[i]);
			}
			HashMap<Integer,Integer> map = new HashMap<>();
			int value = 0;
			for(int a : set)
			{
				map.put(a, value);
				++value;
			}
			Arrays.fill(seg, 0);
			update(1, 0, value, map.get(arr[0]), 1);
			long ans = 0;
		    for(int i = 1;i<n;i++)
		    {
		    	int v1 = query(1,0,value,0,map.get(arr[i])-1);
		    	int v2 = query(1,0,value,map.get(arr[i])+1,value+1);
		    	if(v1<=v2)
		    	{
		    		ans+=v1;
		    	}
		    	else 
		    	{
		    		ans+=v2;
		    	}
		    	update(1,0,value,map.get(arr[i]),1);
		    	
		    }
			sb.append(ans+"\n");
			
			
			

		}
		System.out.print(sb);
	}
    static int seg[] = new int[900000];
    static void update(int si,int ss ,int se ,int x,int value)
    {
    	if(ss == se)
    	{
    		seg[si]+=value;
    		return ;
    	}
    	int mid = (ss+se)/2;
        if(x<=mid)
        	update(2*si,ss,mid,x,value);
        else 
        	update(2*si+1,mid+1,se,x,value);
        seg[si] = seg[2*si]+seg[2*si+1];
    }
    static int query(int si ,int ss ,int se ,int qs,int qe)
    {
    	if(qs<=ss && qe>=se)
    		return seg[si];
    	if(qs>se || ss>qe)
    		return 0;
    	
    	int mid = (ss + se) / 2;
    	return query(2*si,ss,mid,qs,qe) + query(2*si+1,mid+1,se,qs,qe);
    }
	static long gcd(long a, long b) {
		if (a == 0)
			return b;
		else
			return gcd(b % a, a);
	}

	static long lcm(long a, long b) {
		return (a / gcd(a, b)) * b;
	}

	static void sort(long[] a) {
		ArrayList<Long> l = new ArrayList<>();
		for (long i : a)
			l.add(i);
		Collections.sort(l);
		for (int i = 0; i < a.length; i++)
			a[i] = l.get(i);
	}
}

class FastReader {

	byte[] buf = new byte[2048];
	int index, total;
	InputStream in;

	FastReader(InputStream is) {
		in = is;
	}

	int scan() throws IOException {
		if (index >= total) {
			index = 0;
			total = in.read(buf);
			if (total <= 0) {
				return -1;
			}
		}
		return buf[index++];
	}

	String next() throws IOException {
		int c;
		for (c = scan(); c <= 32; c = scan())
			;
		StringBuilder sb = new StringBuilder();
		for (; c > 32; c = scan()) {
			sb.append((char) c);
		}
		return sb.toString();
	}

	String nextLine() throws IOException {
		int c;
		for (c = scan(); c <= 32; c = scan())
			;
		StringBuilder sb = new StringBuilder();
		for (; c != 10 && c != 13; c = scan()) {
			sb.append((char) c);
		}
		return sb.toString();
	}

	char nextChar() throws IOException {
		int c;
		for (c = scan(); c <= 32; c = scan())
			;
		return (char) c;
	}

	int nextInt() throws IOException {
		int c, val = 0;
		for (c = scan(); c <= 32; c = scan())
			;
		boolean neg = c == '-';
		if (c == '-' || c == '+') {
			c = scan();
		}
		for (; c >= '0' && c <= '9'; c = scan()) {
			val = (val << 3) + (val << 1) + (c & 15);
		}
		return neg ? -val : val;
	}

	long nextLong() throws IOException {
		int c;
		long val = 0;
		for (c = scan(); c <= 32; c = scan())
			;
		boolean neg = c == '-';
		if (c == '-' || c == '+') {
			c = scan();
		}
		for (; c >= '0' && c <= '9'; c = scan()) {
			val = (val << 3) + (val << 1) + (c & 15);
		}
		return neg ? -val : val;
	}
}