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.
// package CF_Training.rating_2000;

import java.io.PrintWriter;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class ArrayOptimizationbyDeque {
    //    MUST SEE BEFORE SUBMISSION
//    check whether int part would overflow or not, especially when it is a * b!!!!
//    check if top down dp would cause overflow or not !!!!!!!!!!!!!!!!!!!!!!
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        int t = 1;
        t = sc.nextInt();
        for (int i = 0; i < t; i++) {
            solve(sc, pw);
        }
        pw.close();
    }

    static void solve(Scanner in, PrintWriter out){
        int n = in.nextInt();
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = in.nextInt();
        }
        Treap tp = new Treap(true);
        int[] lo = new int[n], hi = new int[n];
        Map<Integer, Integer> mp = new HashMap<>();
        for (int i = 0; i < n; i++) {
//            int pre = i;
            int sm = tp.orderOfKey(arr[i]);
            lo[i] = sm;
            hi[i] = i - mp.getOrDefault(arr[i], 0) - sm;
            tp.add(arr[i]);
            mp.put(arr[i], mp.getOrDefault(arr[i], 0) + 1);
        }
        long ans = 0;
        for(int i = 0; i < n; i++){
            ans += Math.min(lo[i], hi[i]);
        }
        out.println(ans);
    }

    //findByOrder
//    orderOfKey is used to find the number of value strictly smaller than this value
//    when multi is false which means you cannot have duplicate value and vice versa
    static class Treap {
        private Node root;
        private boolean multi;

        Treap(boolean multi) {
            this.root = null;
            this.multi = multi;
        }

        int size() {
            return size(root);
        }

        boolean isEmpty() {
            return size(root) == 0;
        }

        boolean contains(long key) {
            return contains(root, key);
        }

        void add(long key) {
            root = add(root, key);
        }

        void remove(long key) {
            root = remove(root, key);
        }

        Long first() {
            Node min = findMin(root);
            return min != null ? min.key : null;
        }

        Long last() {
            Node max = findMax(root);
            return max != null ? max.key : null;
        }

        Long poolFirst() {
            Node min = findMin(root);
            if (min != null) {
                remove(min.key);
                return min.key;
            }
            return null;
        }

        Long poolLast() {
            Node max = findMax(root);
            if (max != null) {
                remove(max.key);
                return max.key;
            }
            return null;
        }

        // min >= key
        Long ceiling(long key) {
            return contains(key) ? key : higher(key);
        }

        // max <= key
        Long floor(long key) {
            return contains(key) ? key : lower(key);
        }

        // min > key
        Long higher(long key) {
            Node min = higher(root, key);
            return min == null ? null : min.key;
        }

        private Node higher(Node cur, long key) {
            if (cur == null)
                return null;

            if (cur.key <= key)
                return higher(cur.right, key);

            // cur.key > key
            Node left = higher(cur.left, key);
            return left == null ? cur : left;
        }

        // max < key
        Long lower(long key) {
            Node max = lower(root, key);
            return max == null ? null : max.key;
        }

        private Node lower(Node cur, long key) {
            if (cur == null)
                return null;

            if (cur.key >= key)
                return lower(cur.left, key);

            // cur.key < key
            Node right = lower(cur.right, key);
            return right == null ? cur : right;
        }

        private class Node {
            int height, size;
            long key;
            Node left, right;

            Node(long key) {
                this.key = key;
                height = size = 1;
                left = right = null;
            }
        }

        private int height(Node cur) {
            return cur == null ? 0 : cur.height;
        }

        private int balanceFactor(Node cur) {
            return height(cur.right) - height(cur.left);
        }

        private int size(Node cur) {
            return cur == null ? 0 : cur.size;
        }

        // fixVertex
        private void fixHeightAndSize(Node cur) {
            cur.height = Math.max(height(cur.left), height(cur.right)) + 1;
            cur.size = size(cur.left) + size(cur.right) + 1;
        }

        private Node rotateRight(Node cur) {
            Node prevLeft = cur.left;
            cur.left = prevLeft.right;
            prevLeft.right = cur;
            fixHeightAndSize(cur);
            fixHeightAndSize(prevLeft);
            return prevLeft;
        }

        private Node rotateLeft(Node cur) {
            Node prevRight = cur.right;
            cur.right = prevRight.left;
            prevRight.left = cur;
            fixHeightAndSize(cur);
            fixHeightAndSize(prevRight);
            return prevRight;
        }

        private Node balance(Node cur) {
            fixHeightAndSize(cur);
            if (balanceFactor(cur) == 2) {
                if (balanceFactor(cur.right) < 0)
                    cur.right = rotateRight(cur.right);
                return rotateLeft(cur);
            }
            if (balanceFactor(cur) == -2) {
                if (balanceFactor(cur.left) > 0)
                    cur.left = rotateLeft(cur.left);
                return rotateRight(cur);
            }
            return cur;
        }

        private boolean contains(Node cur, long key) {
            if (cur == null)
                return false;
            else if (key < cur.key)
                return contains(cur.left, key);
            else if (key > cur.key)
                return contains(cur.right, key);
            else
                return true;
        }

        private Node add(Node cur, long key) {
            if (cur == null)
                return new Node(key);

            if (key < cur.key)
                cur.left = add(cur.left, key);
            else if (key > cur.key || multi)
                cur.right = add(cur.right, key);

            return balance(cur);
        }

        private Node findMin(Node cur) {
            return cur.left != null ? findMin(cur.left) : cur;
        }

        private Node findMax(Node cur) {
            return cur.right != null ? findMax(cur.right) : cur;
        }

        private Node removeMin(Node cur) {
            if (cur.left == null)
                return cur.right;

            cur.left = removeMin(cur.left);
            return balance(cur);
        }

        private Node removeMax(Node cur) {
            if (cur.right == null)
                return cur.left;

            cur.right = removeMax(cur.right);
            return balance(cur);
        }

        private Node remove(Node cur, long key) {
            if (cur == null)
                return null;

            if (key < cur.key)
                cur.left = remove(cur.left, key);
            else if (key > cur.key)
                cur.right = remove(cur.right, key);
            else { //  k == cur.key
                Node prevLeft = cur.left;
                Node prevRight = cur.right;

                if (prevRight == null)
                    return prevLeft;

                Node min = findMin(prevRight);
                min.right = removeMin(prevRight);
                min.left = prevLeft;

                return balance(min);
            }

            return balance(cur);
        }

        int orderOfKey(long key) {
            return orderOfKey(root, key);
        }

        // count < key
        private int orderOfKey(Node cur, long key) {
            if (cur == null)
                return 0;

            if (cur.key < key)
                return size(cur.left) + 1 + orderOfKey(cur.right, key);

            if (cur.key == key)
                return size(cur.left);

            // cur.key > key
            return orderOfKey(cur.left, key);
        }

        Long findByOrder(int pos) {
            return size(root) > pos ? findByOrder(root, pos) : null;
        }

        // get i-th
        private long findByOrder(Node cur, int pos) {
            if (size(cur.left) > pos)
                return findByOrder(cur.left, pos);

            if (size(cur.left) == pos)
                return cur.key;

            // size(cur.left) < pos
            return findByOrder(cur.right, pos - 1 - size(cur.left));
        }
    }

}
