Problem: This is an easy version of the problem. In this version, all numbers in the given array are distinct and the constraints on n are less than in the hard version of the problem.

You are given an array a of n integers (there are no equals elements in the array). You can perform the following operations on array elements:

  1. choose any index i (1 ≤ i ≤ n) and move the element a[i] to the begin of the array; 
  2. choose any index i (1 ≤ i ≤ n) and move the element a[i] to the end of the array. 



For example, if n = 5, a = [4, 7, 2, 3, 9], then the following sequence of operations can be performed: 

  * after performing the operation of the first type to the second element, the array a will become [7, 4, 2, 3, 9]; 
  * after performing the operation of the second type to the second element, the array a will become [7, 2, 3, 9, 4]. 



You can perform operations of any type any number of times in any order.

Find the minimum total number of operations of the first and second type that will make the a array sorted in non-decreasing order. In other words, what is the minimum number of operations that must be performed so the array satisfies the inequalities a[1] ≤ a[2] ≤ … ≤ a[n].

Input

The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the test. Then t test cases follow.

Each test case starts with a line containing an integer n (1 ≤ n ≤ 3000) — length of the array a.

Then follow n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9) — an array that needs to be sorted by the given operations. All numbers in the given array are distinct.

The sum of n for all test cases in one test does not exceed 3000.

Output

For each test case output one integer — the minimum total number of operations of the first and second type, which will make the array sorted in non-decreasing order.

Example

Input


4
5
4 7 2 3 9
5
3 5 8 1 7
5
1 4 5 7 12
4
0 2 1 3


Output


2
2
0
2

Note

In the first test case, you first need to move 3, and then 2 to the beginning of the array. Therefore, the desired sequence of operations: [4, 7, 2, 3, 9] → [3, 4, 7, 2, 9] → [2, 3, 4, 7, 9].

In the second test case, you need to move the 1 to the beginning of the array, and the 8 — to the end. Therefore, the desired sequence of operations: [3, 5, 8, 1, 7] → [1, 3, 5, 8, 7] → [1, 3, 5, 7, 8].

In the third test case, the array is already sorted.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class Main {

    public static void main(String[] args) throws Exception {

        new Main().go();
    }

    PrintWriter out;
    Reader in;
    BufferedReader br;

    Main() throws IOException {

        try {

            //br = new BufferedReader( new FileReader("input.txt") );
            //in = new Reader("input.txt");
            in = new Reader("input.txt");
            out = new PrintWriter( new BufferedWriter(new FileWriter("output.txt")) );
        }
        catch (Exception e) {

            //br = new BufferedReader( new InputStreamReader( System.in ) );
            in = new Reader();
            out = new PrintWriter( new BufferedWriter(new OutputStreamWriter(System.out)) );
        }
    }

    void go() throws Exception {

        //int t = in.nextInt();
        int t = 1;
        while (t > 0) {
            solve();
            t--;
        }

        out.flush();
        out.close();
    }


    int inf = 2000000000;
    int mod = 1000000007;
    double eps = 0.000000001;

    int n;
    int m;

    ArrayList<Integer>[] g;
    int[] a;
    void solve() throws IOException {

        int t = in.nextInt();
        while (t > 0) {
            solveCase();
            t--;
        }
    }

    void solveCase() throws IOException {
        int n = in.nextInt();
        int[] a = new int[n];
        int[] b = new int[n + 1];
        for (int i = 0; i < n; i++) {
            a[i] = in.nextInt();
            b[i] = a[i];
        }
        b[n] = -1;
        Arrays.sort(b);
        int best = Integer.MAX_VALUE;
        for (int i = 0; i <= n; i++) {
            int cur = b[i];
            int cnt = 0;
            for (int j = 0; j < n; j++) {
                if (a[j] <= cur) {
                    cnt++;
                }
            }

            int moveMin = Integer.MAX_VALUE;
            int min = Integer.MAX_VALUE;

            int[] needMove = new int[n];

            for (int j = n - 1; j >= 0; j--) {
                if (a[j] <= cur) continue;

                if (a[j] > min) {
                    needMove[j] = 1;
                    moveMin = Math.min(moveMin, a[j]);
                }
                min = Math.min(min, a[j]);
            }

            for (int j = 0; j < n; j++)
                if (a[j] >= moveMin && a[j] > cur) needMove[j] = 1;

            for (int j = 0; j < n; j++) cnt += needMove[j];

            best = Math.min(best, cnt);
        }
        out.println(best);
    }

    static class Pair<A extends Comparable<A>, B extends Comparable<B>> implements Comparable<Pair<A, B>> {

        public final A a;
        public final B b;

        Pair(A a, B b) {
            this.a = a;
            this.b = b;
        }

        public int compareTo(Pair<A, B> p) {
            if (!a.equals(p.a))
                return a.compareTo(p.a);
            else
                return b.compareTo(p.b);
        }
    }

    class Item {

        int a;
        int b;
        int c;

        Item(int a, int b, int c) {
            this.a = a;
            this.b = b;
            this.c = c;
        }

    }


    class Reader {

        BufferedReader  br;
        StringTokenizer tok;

        Reader(String file) throws IOException {
            br = new BufferedReader( new FileReader(file) );
        }

        Reader() throws IOException {
            br = new BufferedReader( new InputStreamReader(System.in) );
        }

        String next() throws IOException {

            while (tok == null || !tok.hasMoreElements())
                tok = new StringTokenizer(br.readLine());
            return tok.nextToken();
        }

        int nextInt() throws NumberFormatException, IOException {
            return Integer.valueOf(next());
        }

        long nextLong() throws NumberFormatException, IOException {
            return Long.valueOf(next());
        }

        double nextDouble() throws NumberFormatException, IOException {
            return Double.valueOf(next());
        }


        String nextLine() throws IOException {
            return br.readLine();
        }

    }

    static class InputReader
    {
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;

        public InputReader()
        {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public InputReader(String file_name) throws IOException
        {
            din = new DataInputStream(new FileInputStream(file_name));
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public String readLine() throws IOException
        {
            byte[] buf = new byte[64]; // line length
            int cnt = 0, c;
            while ((c = read()) != -1)
            {
                if (c == '\n')
                    break;
                buf[cnt++] = (byte) c;
            }
            return new String(buf, 0, cnt);
        }

        public int nextInt() throws IOException
        {
            int ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do
            {
                ret = ret * 10 + c - '0';
            }  while ((c = read()) >= '0' && c <= '9');

            if (neg)
                return -ret;
            return ret;
        }

        public long nextLong() throws IOException
        {
            long ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');
            if (neg)
                return -ret;
            return ret;
        }

        public double nextDouble() throws IOException
        {
            double ret = 0, div = 1;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();

            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');

            if (c == '.')
            {
                while ((c = read()) >= '0' && c <= '9')
                {
                    ret += (c - '0') / (div *= 10);
                }
            }

            if (neg)
                return -ret;
            return ret;
        }

        private void fillBuffer() throws IOException
        {
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
            if (bytesRead == -1)
                buffer[0] = -1;
        }

        private byte read() throws IOException
        {
            if (bufferPointer == bytesRead)
                fillBuffer();
            return buffer[bufferPointer++];
        }

        public void close() throws IOException
        {
            if (din == null)
                return;
            din.close();
        }
    }

}