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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.PrintStream;
import java.util.Iterator;
import java.util.ArrayList;
import java.util.AbstractCollection;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author sarthakmanna
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastReader in = new FastReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        E2ArrayOptimizationByDeque solver = new E2ArrayOptimizationByDeque();
        int testCount = Integer.parseInt(in.next());
        for (int i = 1; i <= testCount; i++)
            solver.solve(i, in, out);
        out.close();
    }

    static class E2ArrayOptimizationByDeque {
        public void solve(int testNumber, FastReader in, PrintWriter out) {
            int i, j, k;

            int N = in.nextInt();
            int[] A = in.getIntArray(N);

            MyTreeSetLong set = new MyTreeSetLong(true);
            long ans = 0;
            for (i = 0; i < N; ++i) {
                ans += Math.min(set.size() - set.countFloorNodeLongs(A[i]),
                        set.countFloorNodeLongs(A[i] - 1));
                set.add(A[i]);
            }
            out.println(ans);
        }

    }

    static class MyTreeSetLong implements Iterable<Long> {
        NodeLong root;
        boolean allowDuplicates;
        private int itrInd;
        private ArrayList<NodeLong> dfsTrav = new ArrayList<>();

        public MyTreeSetLong() {
            root = null;
        }

        public MyTreeSetLong(boolean allowDupli) {
            root = null;
            allowDuplicates = allowDupli;
        }

        public boolean add(long value) {
            return add(new NodeLong(value));
        }

        public int countFloorNodeLongs(long value) {
            return countFloorNodeLongs(root, new NodeLong(value));
        }

        public int size() {
            return root == null ? 0 : root.size();
        }

        private boolean add(NodeLong toAdd) {
            if (root == null) {
                root = toAdd;
                return true;
            }

            NodeLong tr = root;
            while (true) {
                if (toAdd.compareTo(tr) < 0) {
                    if (tr.left == null) {
                        tr.left = toAdd;
                        toAdd.parent = tr;
                        break;
                    } else tr = tr.left;
                } else if (allowDuplicates || toAdd.compareTo(tr) > 0) {
                    if (tr.right == null) {
                        tr.right = toAdd;
                        toAdd.parent = tr;
                        break;
                    } else tr = tr.right;
                } else return false;
            }

            while (toAdd != null) {
                root = toAdd.finalisePosition(root);
                toAdd = toAdd.parent;
            }
            return true;
        }

        private int countFloorNodeLongs(NodeLong node, NodeLong key) {
            if (node == null) return 0;
            else if (node.compareTo(key) > 0)
                return countFloorNodeLongs(node.left, key);
            else
                return NodeLong.findSize(node.left) + 1 + countFloorNodeLongs(node.right, key);
        }

        public Iterator<Long> iterator() {
            dfsTrav.clear();
            dfs(root);
            itrInd = 0;
            Iterator<Long> iterator = new Iterator<Long>() {

                public boolean hasNext() {
                    return itrInd < size();
                }


                public Long next() {
                    return dfsTrav.get(itrInd++).value;
                }
            };
            return iterator;
        }

        private void dfs(NodeLong node) {
            if (node == null) return;
            dfs(node.left);
            dfsTrav.add(node);
            dfs(node.right);
        }

        public String toString() {
            dfsTrav.clear();
            dfs(root);
            return dfsTrav.toString();
        }

    }

    static class FastReader {
        static final int BUFSIZE = 1 << 20;
        static byte[] buf;
        static int index;
        static int total;
        static InputStream in;

        public FastReader(InputStream is) {
            try {
                in = is;
                buf = new byte[BUFSIZE];
            } catch (Exception e) {
            }
        }

        private int scan() {
            try {
                if (index >= total) {
                    index = 0;
                    total = in.read(buf);
                    if (total <= 0)
                        return -1;
                }
                return buf[index++];
            } catch (Exception | Error e) {
                System.err.println(e.getMessage());
                return 13 / 0;
            }
        }

        public String next() {
            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();
        }

        public int nextInt() {
            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;
        }

        public int[] getIntArray(int size) {
            int[] ar = new int[size];
            for (int i = 0; i < size; ++i) ar[i] = nextInt();
            return ar;
        }

    }

    static class NodeLong {
        NodeLong parent;
        NodeLong left;
        NodeLong right;
        long value;
        int size;
        int height;

        NodeLong(long v) {
            value = v;
            size = height = 1;
        }

        NodeLong finalisePosition(NodeLong root) {
            int lheight = findDepth(left), rheight = findDepth(right);

            if (lheight > rheight + 1) {
                if (findDepth(left.right) > findDepth(left.left))
                    root = left.rotateLeft(root);
                root = rotateRight(root);
            } else if (lheight + 1 < rheight) {
                if (findDepth(right.left) > findDepth(right.right))
                    root = right.rotateRight(root);
                root = rotateLeft(root);
            } else {
                setParameters();
            }

            return root;
        }

        NodeLong rotateRight(NodeLong root) {
            NodeLong A = this, B = left, P = parent;

            if (P != null) {
                if (P.left == this) P.left = B;
                else if (P.right == this) P.right = B;
                else System.exit(7 / 0);
            } else root = B;
            B.parent = A.parent;

            A.left = B.right;
            if (B.right != null) B.right.parent = A;
            B.right = A;
            A.parent = B;

            A.setParameters();
            B.setParameters();

            return root;
        }

        NodeLong rotateLeft(NodeLong root) {
            NodeLong A = this, B = right, P = parent;

            if (P != null) {
                if (P.left == this) P.left = B;
                else if (P.right == this) P.right = B;
                else System.exit(7 / 0);
            } else root = B;
            B.parent = A.parent;

            A.right = B.left;
            if (B.left != null) B.left.parent = A;
            B.left = A;
            A.parent = B;

            A.setParameters();
            B.setParameters();

            return root;
        }

        void setParameters() {
            size = 1 + findSize(left) + findSize(right);
            height = 1 + Math.max(findDepth(left), findDepth(right));
        }

        int size() {
            return size;
        }

        static int findDepth(NodeLong node) {
            return node == null ? 0 : node.height;
        }

        static int findSize(NodeLong node) {
            return node == null ? 0 : node.size();
        }

        public int compareTo(NodeLong node) {
            return Long.compare(value, node.value);
        }

        public String toString() {
            return value + "";
        }

    }
}

