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.*;
import java.util.*;
import java.math.BigInteger;

/**
 __                  __
 ( _)                ( _)
 / / \\              / /\_\_
 / /   \\            / / | \ \
 / /     \\          / /  |\ \ \
 /  /   ,  \ ,       / /   /|  \ \
 /  /    |\_ /|      / /   / \   \_\
 /  /  |\/ _ '_| \   / /   /   \    \\
 |  /   |/  0 \0\    / |    |    \    \\
 |    |\|      \_\_ /  /    |     \    \\
 |  | |/    \.\ o\o)  /      \     |    \\
 \    |     /\\`v-v  /        |    |     \\
 | \/    /_| \\_|  /         |    | \    \\
 | |    /__/_ `-` /   _____  |    |  \    \\
 \|    [__]  \_/  |_________  \   |   \    ()
 /    [___] (    \         \  |\ |   |   //
 |    [___]                  |\| \|   /  |/
 /|    [____]                  \  |/\ / / ||
 (  \   [____ /     ) _\      \  \    \| | ||
 \  \  [_____|    / /     __/    \   / / //
 |   \ [_____/   / /        \    |   \/ //
 |   /  '----|   /=\____   _/    |   / //
 __ /  /        |  /   ___/  _/\    \  | ||
 (/-(/-\)       /   \  (/\/\)/  |    /  | /
 (/\/\)           /   /   //
 _________/   /    /
 \____________/    (


 @author NTUDragons-Reborn
 */

public class C{
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Task solver = new Task();
        solver.solve(in, out);
        out.close();
    }
    // main solver
    static class Task{

        double eps= 0.00000001;
        static final int MAXN = 200015;
        static final int MOD= 1000000007;

        // stores smallest prime factor for every number
        static int spf[] = new int[MAXN];

        static boolean[] prime;

        Map<Integer,Set<Integer>> dp= new HashMap<>();

        // Calculating SPF (Smallest Prime Factor) for every
        // number till MAXN.
        // Time Complexity : O(nloglogn)
        public void sieve()
        {
            spf[1] = 1;
            for (int i=2; i<MAXN; i++)

                // marking smallest prime factor for every
                // number to be itself.
                spf[i] = i;

            // separately marking spf for every even
            // number as 2
            for (int i=4; i<MAXN; i+=2)
                spf[i] = 2;

            for (int i=3; i*i<MAXN; i++)
            {
                // checking if i is prime
                if (spf[i] == i)
                {
                    // marking SPF for all numbers divisible by i
                    for (int j=i*i; j<MAXN; j+=i)

                        // marking spf[j] if it is not
                        // previously marked
                        if (spf[j]==j)
                            spf[j] = i;
                }
            }
        }

        void sieveOfEratosthenes(int n)
        {
            // Create a boolean array
            // "prime[0..n]" and
            // initialize all entries
            // it as true. A value in
            // prime[i] will finally be
            // false if i is Not a
            // prime, else true.
            prime= new boolean[n + 1];
            for (int i = 0; i <= n; i++)
                prime[i] = true;
    
            for (int p = 2; p * p <= n; p++)
            {
                // If prime[p] is not changed, then it is a
                // prime
                if (prime[p] == true)
                {
                    // Update all multiples of p
                    for (int i = p * p; i <= n; i += p)
                        prime[i] = false;
                }
            }
        }

        // A O(log n) function returning primefactorization
        // by dividing by smallest prime factor at every step
        public Set<Integer> getFactorization(int x)
        {
            if(dp.containsKey(x)) return dp.get(x);
            Set<Integer> ret = new HashSet<>();
            while (x != 1)
            {
                if(spf[x]!=2) ret.add(spf[x]);
                x = x / spf[x];
            }
            dp.put(x,ret);
            return ret;
        }
        // function to find first index >= x
        public int lowerIndex(List<Integer> arr, int n, int x)
        {
            int l = 0, h = n - 1;
            while (l <= h)
            {
                int mid = (l + h) / 2;
                if (arr.get(mid) >= x)
                    h = mid - 1;
                else
                    l = mid + 1;
            }
            return l;
        }
        public int lowerIndex(int[] arr, int n, int x)
        {
            int l = 0, h = n - 1;
            while (l <= h)
            {
                int mid = (l + h) / 2;
                if (arr[mid] >= x)
                    h = mid - 1;
                else
                    l = mid + 1;
            }
            return l;
        }

        // function to find last index <= y
        public int upperIndex(List<Integer> arr, int n, int y)
        {
            int l = 0, h = n - 1;
            while (l <= h)
            {
                int mid = (l + h) / 2;
                if (arr.get(mid) <= y)
                    l = mid + 1;
                else
                    h = mid - 1;
            }
            return h;
        }
        public int upperIndex(int[] arr, int n, int y)
        {
            int l = 0, h = n - 1;
            while (l <= h)
            {
                int mid = (l + h) / 2;
                if (arr[mid] <= y)
                    l = mid + 1;
                else
                    h = mid - 1;
            }
            return h;
        }

        // function to count elements within given range
        public int countInRange(List<Integer> arr, int n, int x, int y)
        {
            // initialize result
            int count = 0;
            count = upperIndex(arr, n, y) -
                    lowerIndex(arr, n, x) + 1;
            return count;
        }

        public int add(int a, int b){
            a+=b;
            if(a>=MOD) a-=MOD;
            else if(a<0) a+=MOD;
            return a;
        }

        public int mul(int a, int b){
            long res= (long)a*(long)b;
            return (int)(res%MOD);
        }

        public int power(int a, int b) {
            int ans=1;
            while(b>0){
                if((b&1)!=0) ans= mul(ans,a);
                b>>=1;
                a= mul(a,a);
            }
            return ans;
        }

        int[] fact= new int[MAXN];
        int[] inv= new int[MAXN];

        public int Ckn(int n, int k){
            if(k<0 || n<0) return 0;
            return mul(mul(fact[n],inv[k]),inv[n-k]);
        }

        public int inverse(int a){
            return power(a,MOD-2);
        }

        public void preprocess() {
            fact[0]=1;
            for(int i=1;i<MAXN;i++) fact[i]= mul(fact[i-1],i);
            inv[MAXN-1]= inverse(fact[MAXN-1]);
            for(int i=MAXN-2;i>=0;i--){
                inv[i]= mul(inv[i+1],i+1);
            }
        }

        /**
         * return VALUE of lower bound for unsorted array
         */
        public int lowerBoundNormalArray(int[] arr, int x){
            TreeSet<Integer> set= new TreeSet<>();
            for(int num: arr) set.add(num);
            return set.lower(x);
        }
        /**
         * return VALUE of upper bound for unsorted array
         */
        public int upperBoundNormalArray(int[] arr, int x){
            TreeSet<Integer> set= new TreeSet<>();
            for(int num: arr) set.add(num);
            return set.higher(x);
        }

        public void debugArr(int[] arr){
            for(int i: arr) out.print(i+" ");
            out.println();
        }

        public int rand(){
            int min=0, max= MAXN;
            int random_int = (int)Math.floor(Math.random()*(max-min+1)+min);
            return random_int;
        }
        
        InputReader in; PrintWriter out;
        CustomFileReader cin;
        int[] xor= new int[3*100000+5];
        public void solve(InputReader in, PrintWriter out) {
            this.in=in; this.out=out;
            // cin= new CustomFileReader("../HackerCup21//realA.txt");
            // preprocess();
            int t=in.nextInt();
            // int t= cin.nextIntArrLine()[0]; 
            for(int i=1;i<=t;i++) solveA(i);
        }
        public void solveA(int t){
            int n= in.nextInt();
            int[] a= new int[n];
            for(int i=0;i<n;i++) a[i]= in.nextInt();
            Set<Integer> set= new HashSet<>();
            List<Integer> list= new ArrayList<>();
            for(int i: a) set.add(i);
            for(int i: set) list.add(i);
            Collections.sort(list);
            Map<Integer,Integer> idx= new HashMap<>();
            int index=0;
            for(int i: list){
                idx.put(i,index);
                index++;
            }
            long res=0;
            BIT fenwick= new BIT();
            for(int i=0;i<n;i++){
                index= idx.get(a[i]);
                res+= Math.min(fenwick.getSum(index-1) , i-fenwick.getSum(index));

                fenwick.updateBIT(n, index, 1);
            }
            out.println(res);

        }
        public void solveB(int t) {
        }
        public void solveC(int t){
        
        }
        public int _gcd(int a,  int b)
        {

            if(b == 0) {
                return a;
            }
            else {
                return _gcd(b, a % b);
            }
        }
        public int _lcm(int a, int b){
            return (a*b)/_gcd(a,b);
        }

    }


    static class SEG {
        public long _gcd(long a,  long  b)
        {

            if(b == 0) {
                return a;
            }
            else {
                return _gcd(b, a % b);
            }
        }
        long[] segs;
        public SEG(List<Long> a) {
            int n= a.size();
            segs= new long[4*n];
            build(0,a,0,n-1);
        }
        void build(int root, List<Long> a, int l, int r){
            if (l==r){
                segs[root] = a.get(l); return;
            }
            int m = (l+r) >> 1;
            build(root*2+1, a, l, m);
            build(root*2+2, a, m+1, r);
            segs[root] = _gcd(segs[root*2+1], segs[root*2+2]);
        }
        long query(int root, int l, int r, int lq, int rq){
            if (lq > r || rq < l) return 0l;
            if (l >= lq && rq >= r) return segs[root];
            int m = (l+r)>>1;
            return _gcd(query(root*2+1, l, m, lq, rq),
                         query(root*2+2, m+1, r, lq, rq));
        }
    }

    static class BIT {
        int MAXN= 200015;
        int BITree[] = new int[MAXN];
     
        /* n --> No. of elements present in input array.
        BITree[0..n] --> Array that represents Binary
                        Indexed Tree.
        arr[0..n-1] --> Input array for which prefix sum
                        is evaluated. */
    
        // Returns sum of arr[0..index]. This function
        // assumes that the array is preprocessed and
        // partial sums of array elements are stored
        // in BITree[].
        public int getSum(int index)
        {
            int sum = 0; // Initialize result
        
            // index in BITree[] is 1 more than
            // the index in arr[]
            index = index + 1;
        
            // Traverse ancestors of BITree[index]
            while(index>0)
            {
                // Add current element of BITree
                // to sum
                sum += BITree[index];
        
                // Move index to parent node in
                // getSum View
                index -= index & (-index);
            }
            return sum;
        }
    
        // Updates a node in Binary Index Tree (BITree)
        // at given index in BITree. The given value
        // 'val' is added to BITree[i] and all of
        // its ancestors in tree.
        public void updateBIT(int n, int index,
                                            int val)
        {
            // index in BITree[] is 1 more than
            // the index in arr[]
            index = index + 1;
        
            // Traverse all ancestors and add 'val'
            while(index <= n)
            {
            // Add 'val' to current node of BIT Tree
            BITree[index] += val;
        
            // Update index to that of parent
            // in update View
            index += index & (-index);
            }
        }
    
        /* Function to construct fenwick tree
        from given array.*/
        void constructBITree(int arr[], int n)
        {
            // Initialize BITree[] as 0
            for(int i=1; i<=n; i++)
                BITree[i] = 0;
        
            // Store the actual values in BITree[]
            // using update()
            for(int i = 0; i < n; i++)
                updateBIT(n, i, arr[i]);
        }
    }

    static class Venice{
        public Map<Long,Long> m= new HashMap<>();
        public long base=0;
        public long totalValue=0;
        private int M= 1000000007;

        private long addMod(long a, long b){
            a+=b;
            if(a>=M) a-=M;
            return a;
        }

        public void reset(){
            m= new HashMap<>();
            base=0;
            totalValue=0;
        }
        public void update(long add){
            base= base+ add;
        }
        public void add(long key, long val){
            long newKey= key-base;
            m.put(newKey, addMod(m.getOrDefault(newKey,(long)0),val));
        }
    }



    static class Tuple implements Comparable<Tuple>{
        int x, y, z;
        public Tuple(int x, int y, int z){
            this.x= x;
            this.y= y;
            this.z=z;
        }
        @Override
        public int compareTo(Tuple o){
            if(this.x!=o.x) return this.x-o.x;
            if(this.y!=o.y) return this.y-o.y;
            return this.z-o.z;
        }
    }

    static class Pair implements Comparable<Pair>{
        public int x;
        public int y;
        public Pair(int x, int y){
            this.x= x;
            this.y= y;
        }

        @Override
        public int compareTo(Pair o) {
            return this.x-o.x;
        }
    }

    // public static class compareL implements Comparator<Tuple>{
    //     @Override
    //     public int compare(Tuple t1, Tuple t2) {
    //         return t2.l - t1.l;
    //     }
    // }

    // fast input reader class;
    static class InputReader {
        BufferedReader br;
        StringTokenizer st;

        public InputReader(InputStream stream) {
            br = new BufferedReader(new InputStreamReader(stream));
        }

        public String nextToken() {
            while (st == null || !st.hasMoreTokens()) {
                String line = null;
                try {
                    line = br.readLine();
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
                if (line == null) {
                    return null;
                }
                st = new StringTokenizer(line);
            }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(nextToken());
        }
        public double nextDouble(){
            return Double.parseDouble(nextToken());
        }
        public long nextLong(){
            return Long.parseLong(nextToken());
        }
        public int[] nextIntArr(int n){
            int[] arr= new int[n];
            for(int i=0;i<n;i++) arr[i]= nextInt();
            return arr;
        }
        public long[] nextLongArr(int n){
            long[] arr= new long[n];
            for(int i=0;i<n;i++) arr[i]= nextLong();
            return arr;
        }
        public List<Integer> nextIntList(int n){
            List<Integer> arr= new ArrayList<>();
            for(int i=0;i<n;i++) arr.add(nextInt());
            return arr;
        }
        public int[][] nextIntMatArr(int n, int m){
            int[][] mat= new int[n][m];
            for(int i=0;i<n;i++) for(int j=0;j<m;j++) mat[i][j]= nextInt();
            return mat;
        }
        public List<List<Integer>> nextIntMatList(int n, int m){
            List<List<Integer>> mat= new ArrayList<>();
            for(int i=0;i<n;i++){
                List<Integer> temp= new ArrayList<>();
                for(int j=0;j<m;j++) temp.add(nextInt());
                mat.add(temp);
            }
            return mat;
        }
        public char[] nextStringCharArr(){
            return nextToken().toCharArray();
        }
    }

    static class CustomFileReader{
        String path="";
        Scanner sc;
        public CustomFileReader(String path){
            this.path=path;
            try{
                sc= new Scanner(new File(path));
            }
            catch(Exception e){}
        }
        public String nextLine(){
            return sc.nextLine();
        }
        public int[] nextIntArrLine(){
            String line= sc.nextLine();
            String[] part= line.split("[\\s+]");
            int[] res= new int[part.length];
            for(int i=0;i<res.length;i++) res[i]= Integer.parseInt(part[i]);
            return res;
        }
    }
}