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 F {

    void solve() throws IOException {
        in = new InputReader("__std");
        out = new OutputWriter("__std");

        int testCount = in.readInt();
        for (int test = 1; test <= testCount; ++test) {
            int n = in.readInt();
            int[] a = new int[n];
            Map<Integer, List<Integer>> pos = new HashMap<>();
            for (int i = 0; i < n; ++i) {
                a[i] = in.readInt();
                if (!pos.containsKey(a[i])) {
                    pos.put(a[i], new ArrayList<>());
                }
                pos.get(a[i]).add(i);
            }
            Arrays.sort(a);
            int best = 0;
            int cur = pos.get(a[0]).size();
            for (int i = 1; i < n; ++i) {
                if (a[i] != a[i - 1]) {
                    List<Integer> u = pos.get(a[i - 1]);
                    List<Integer> v = pos.get(a[i]);
                    // <= 2
                    int l = 0;
                    for (int r = 0; r < v.size(); ++r) {
                        while (l < u.size() && u.get(l) < v.get(r)) {
                            ++l;
                        }
                        best = Math.max(best, l + v.size() - r);
                    }
                    // >= 3
                    if (u.get(u.size() - 1) < v.get(0)) {
                        cur += v.size();
                    } else {
                        best = Math.max(best, cur + v.size() + Collections.binarySearch(v, u.get(u.size() - 1)) + 1);
                        cur = v.size() - Collections.binarySearch(u, v.get(0)) - 1;
                    }
                }
            }
            best = Math.max(best, cur);
            out.println(n - best);
        }

        exit();
    }

    void exit() {
        out.close();
        //System.err.println((System.currentTimeMillis() - startTime) + " ms");
        System.exit(0);
    }

    InputReader in;
    OutputWriter out;

    //long startTime = System.currentTimeMillis();

    public static void main(String[] args) throws IOException {
        new F().solve();
    }

    class InputReader {

        private InputStream stream;

        private byte[] buffer = new byte[1024];
        private int pos, len;

        private int cur;

        private StringBuilder sb = new StringBuilder(32);

        InputReader(String name) throws IOException {
            if (name.equals("__std")) {
                stream = System.in;
            } else {
                stream = new FileInputStream(name);
            }
            cur = read();
        }

        private int read() throws IOException {
            if (len == -1) {
                throw new EOFException();
            }
            if (pos >= len) {
                pos = 0;
                len = stream.read(buffer);
                if (len == -1) return -1;
            }
            return buffer[pos++] & 0xff;
        }

        boolean whitespace() {
            return cur == ' ' || cur == '\t' || cur == '\r' || cur == '\n' || cur == -1;
        }

        boolean eol() {
            return cur == '\r' || cur == '\n' || cur == -1;
        }

        boolean eof() {
            return cur == -1;
        }

        byte readByte() throws IOException {
            if (eof()) {
                throw new EOFException();
            }
            byte res = (byte) cur;
            cur = read();
            return res;
        }

        char readChar() throws IOException {
            if (eof()) {
                throw new EOFException();
            }
            char res = (char) cur;
            cur = read();
            return res;
        }

        int readInt() throws IOException {
            if (eof()) {
                throw new EOFException();
            }
            while (whitespace()) {
                cur = read();
            }
            if (eof()) {
                throw new EOFException();
            }
            int sign = 1;
            if (cur == '-') {
                sign = -1;
                cur = read();
                if (cur < '0' || cur > '9') {
                    throw new NumberFormatException();
                }
            }
            int res = 0;
            while (!whitespace()) {
                if (cur < '0' || cur > '9') {
                    throw new NumberFormatException();
                }
                res *= 10;
                res += cur - '0';
                cur = read();
            }
            return res * sign;
        }

        long readLong() throws IOException {
            if (eof()) {
                throw new EOFException();
            }
            while (whitespace()) {
                cur = read();
            }
            if (eof()) {
                throw new EOFException();
            }
            int sign = 1;
            if (cur == '-') {
                sign = -1;
                cur = read();
                if (cur < '0' || cur > '9') {
                    throw new NumberFormatException();
                }
            }
            long res = 0;
            while (!whitespace()) {
                if (cur < '0' || cur > '9') {
                    throw new NumberFormatException();
                }
                res *= 10;
                res += cur - '0';
                cur = read();
            }
            return res * sign;
        }

        double readDouble() throws IOException {
            return Double.parseDouble(readToken());
        }

        String readToken() throws IOException {
            if (eof()) {
                throw new EOFException();
            }
            while (whitespace()) {
                cur = read();
            }
            if (eof()) {
                throw new EOFException();
            }
            sb.setLength(0);
            while (!whitespace()) {
                sb.append((char) cur);
                cur = read();
            }
            return sb.toString();
        }

        String readLine() throws IOException {
            if (eof()) {
                throw new EOFException();
            }
            sb.setLength(0);
            while (!eol()) {
                sb.append((char) cur);
                cur = read();
            }
            if (cur == '\r') {
                cur = read();
            }
            if (cur == '\n') {
                cur = read();
            }
            return sb.toString();
        }

        void skipLine() throws IOException {
            if (eof()) {
                throw new EOFException();
            }
            while (!eol()) {
                cur = read();
            }
            if (cur == '\r') {
                cur = read();
            }
            if (cur == '\n') {
                cur = read();
            }
        }
    }

    class OutputWriter {

        private PrintWriter writer;

        OutputWriter(String name) throws IOException {
            if (name.equals("__std")) {
                writer = new PrintWriter(System.out);
            } else {
                writer = new PrintWriter(name);
            }
        }

        void print(String format, Object ... args) {
            writer.print(new Formatter(Locale.US).format(format, args));
        }

        void println(String format, Object ... args) {
            writer.println(new Formatter(Locale.US).format(format, args));
        }

        void print(Object value) {
            writer.print(value);
        }

        void println(Object value) {
            writer.println(value);
        }

        void println() {
            writer.println();
        }

        void printArray(int[] a) {
            boolean first = true;
            for (int v : a) {
                if (!first) {
                    writer.print(' ');
                } else {
                    first = false;
                }
                writer.print(v);
            }
            writer.println();
        }

        void printArray(int[][] a) {
            for (int[] v : a) {
                printArray(v);
            }
        }

        void flush() {
            writer.flush();
        }

        void close() {
            writer.close();
        }
    }
}