description: 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.
solution:  //package eround101;
  
  import java.util.*;
  import java.io.*;
  import java.lang.*;
  import java.util.StringTokenizer;
  
  public class C101 {
  
      static HritikScanner sc = new HritikScanner();
      static PrintWriter pw = new PrintWriter(System.out, true);
      final static int MOD = (int) 1e9 + 7;
      static StringBuilder sb = new StringBuilder();
  
      public static void main(String[] args) {
  
          int t = ni();
          while (t-- > 0) {
              solve();
          }
      }
  
      static void solve() {
          int n = ni();
          int[] arr = nextIntArray(n);
          ArrayList<Integer> l = new ArrayList<>();
          for(int ele : arr)
          {
              l.add(ele);
          }
          Collections.sort(l);
          Map<Integer, Integer> map = new HashMap<>();
          int ind = 1;
          for(int ele : l)
          {
              if(!map.containsKey(ele))
              {
                  map.put(ele, ind++);
              }
          }
          int[] comp = new int[n];
          for(int i = 0; i < n; i++)
          {
              comp[i] = map.get(arr[i]);
          }
          
          FenwickTree ft = new FenwickTree(n);
          long ans = 0;
          for(int i = 0; i < n; i++)
          {
              int compVal = comp[i];
              int small = ft.sum(1, compVal-1);
              int big = ft.sum(compVal+1, n);
              ans += Math.min(small, big);
              ft.add(compVal, 1);// we are storing the frequency not the  value itself.
          }
          pl(ans);
          
      }
      
      /////////////////////////////////////////////////////////////////////////////////
      static class FenwickTree{ // Binary Index Tree
          int[] tree;
          static int size;
          
          public FenwickTree(int size)
          {
              this.size = size;
              tree = new int[size+5];
          }
          
          public void add(int i, int val)
          {
              while(i <= size)
              {
                  tree[i] += val;
                  i += i & -i;  // adding the decimal value of the last set bit.
              }
          }
          
          public int sum(int i)
          {
              int res = 0;
              while(i >= 1)
              {
                  res += tree[i];
                  i -= i & -i; // deleting the last set bit
              }
              return res;
          }
          
          public int sum(int l, int r)
          {
              return sum(r) - sum(l-1);
          }
      }
      
      /////////////////////////////////////////////////////////////////////////////////
      static int[] nextIntArray(int n) {
          int[] arr = new int[n];
          int i = 0;
          while (i < n) {
              arr[i++] = ni();
          }
          return arr;
      }
  
      static long[] nextLongArray(int n) {
          long[] arr = new long[n];
          int i = 0;
          while (i < n) {
              arr[i++] = nl();
          }
          return arr;
      }
  
      static int[] nextIntArray1(int n) {
          int[] arr = new int[n + 1];
          int i = 1;
          while (i <= n) {
              arr[i++] = ni();
          }
          return arr;
      }
  
      /////////////////////////////////////////////////////////////////////////////////
      static int ni() {
          return sc.nextInt();
      }
  
      static long nl() {
          return sc.nextLong();
      }
  
      static double nd() {
          return sc.nextDouble();
      }
  
      /////////////////////////////////////////////////////////////////////////////////
      static void pl() {
          pw.println();
      }
  
      static void p(Object o) {
          pw.print(o+ " ");
      }
  
      static void pl(Object o) {
          pw.println(o);
      }
  
      static void psb(StringBuilder sb) {
          pw.print(sb);
      }
  
      static void pa(Object arr[]) {
          for (Object o : arr) {
              p(o);
          }
          pl();
      }
  
      static void pa(int arr[]) {
          for (int o : arr) {
              p(o);
          }
          pl();
      }
  
      static void pa(long arr[]) {
          for (long o : arr) {
              p(o);
          }
          pl();
      }
  
      static void pa(double arr[]) {
          for (double o : arr) {
              p(o);
          }
          pl();
      }
  
      static void pa(char arr[]) {
          for (char o : arr) {
              p(o);
          }
          pl();
      }
  
      static void pa(List list) {
          for (Object o : list) {
              p(o);
          }
          pl();
      }
  
      static void pa(Object[][] arr) {
  
          for (int i = 0; i < arr.length; ++i) {
              for (Object o : arr[i]) {
                  p(o);
              }
              pl();
          }
      }
  
      static void pa(int[][] arr) {
          for (int i = 0; i < arr.length; ++i) {
              for (int o : arr[i]) {
                  p(o);
              }
              pl();
          }
      }
  
      static void pa(long[][] arr) {
          for (int i = 0; i < arr.length; ++i) {
              for (long o : arr[i]) {
                  p(o);
              }
              pl();
          }
      }
  
      static void pa(char[][] arr) {
          for (int i = 0; i < arr.length; ++i) {
              for (char o : arr[i]) {
                  p(o);
              }
              pl();
          }
      }
  
      static void pa(double[][] arr) {
          for (int i = 0; i < arr.length; ++i) {
              for (double o : arr[i]) {
                  p(o);
              }
              pl();
          }
      }
  
  /////////////////////////////////////////////////////////////////////////////////
      static void print(Object s) {
          System.out.println(s);
      }
  
      /////////////////////////////////////////////////////////////////////////////////
      //-----------HritikScanner class for faster input----------//
      static class HritikScanner {
  
          BufferedReader br;
          StringTokenizer st;
  
          public HritikScanner() {
              br = new BufferedReader(new InputStreamReader(System.in));
          }
  
          String next() {
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.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 e) {
                  e.printStackTrace();
              }
              return str;
          }
      }
  
      //////////////////////////////////////////////////////////////////
      public static class Pair implements Comparable<Pair> {
  
          int num, row,col;
  
          Pair(int num, int row, int col) {
              this.num = num;
              this.row = row;
              this.col = col;
          }
  
          public int get1() {
              return num;
          }
  
          public int get2() {
              return row;
          }
  
          public int compareTo(Pair A) {
              return this.num - A.num;
          }
          
          public String toString()
          {
              return num +" "+ row+" "+col;
          }
      }
  
      //////////////////////////////////////////////////////////////////
      // Function to return gcd of a and b time complexity O(log(a+b))
      static long gcd(long a, long b) {
          if (a == 0) {
              return b;
          }
          return gcd(b % a, a);
      }
  
      // method to return LCM of two numbers 
      static long lcm(int a, int b) {
          return (a / gcd(a, b)) * b;
      }
  
      //////////////////////////////////////////////////////////////////
      static boolean isPrime(long n) {
          // Corner cases 
          if (n <= 1) {
              return false;
          }
          if (n <= 3) {
              return true;
          }
  
          // This is checked so that we can skip  
          // middle five numbers in below loop 
          if (n % 2 == 0 || n % 3 == 0) {
              return false;
          }
  
          for (long i = 5; i * i <= n; i = i + 6) {
              if (n % i == 0 || n % (i + 2) == 0) {
                  return false;
              }
          }
  
          return true;
      }
      //////////////////////////////////////////////////////////////////
  
      static boolean isPowerOfTwo(long n) {
          if (n == 0) {
              return false;
          }
  
          return (long) (Math.ceil((Math.log(n) / Math.log(2))))
                  == (long) (Math.floor(((Math.log(n) / Math.log(2)))));
      }
  
      public static long getFact(int n) {
          long ans = 1;
          while (n > 0) {
              ans *= n;
              ans %= MOD;
              n--;
          }
          return ans;
      }
  
      public static long pow(long n, int pow) {
          if (pow == 0) {
              return 1;
          }
          long temp = pow(n, pow / 2) % MOD;
          temp *= temp;
          temp %= MOD;
          if (pow % 2 == 1) {
              temp *= n;
          }
          temp %= MOD;
          return temp;
  
      }
  
      public static long nCr(int n, int r) {
          long ans = 1;
          int temp = n - r;
          while (n > temp) {
              ans *= n;
              ans %= MOD;
              n--;
          }
          ans *= pow(getFact(r) % MOD, MOD - 2) % MOD;
          ans %= MOD;
          return ans;
      }
  
      //////////////////////////////////////////////////////////////////
      // method returns Nth power of A 
      static double nthRoot(int A, int N) {
  
          // intially guessing a random number between 
          // 0 and 9 
          double xPre = Math.random() % 10;
  
          // smaller eps, denotes more accuracy 
          double eps = 0.001;
  
          // initializing difference between two 
          // roots by INT_MAX 
          double delX = 2147483647;
  
          // xK denotes current value of x 
          double xK = 0.0;
  
          // loop untill we reach desired accuracy 
          while (delX > eps) {
              // calculating current value from previous 
              // value by newton's method 
              xK = ((N - 1.0) * xPre
                      + (double) A / Math.pow(xPre, N - 1)) / (double) N;
              delX = Math.abs(xK - xPre);
              xPre = xK;
          }
  
          return xK;
      }
  
  }
  