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.util.*;
import java.io.*;
//Captain on duty!



public class Main {

    static void compare(Main.pair a[], int n) {

        Arrays.sort(a, new Comparator<Main.pair>() {
            @Override
            public int compare(Main.pair p1, Main.pair p2) {
                return p1.f - p2.f;
            }
        });
    }

    public static boolean checkPalindrome(String s) {
        // reverse the given String
        String reverse = new StringBuffer(s).reverse().toString();

        // check whether the string is palindrome or not
        if (s.equals(reverse))
            return true;

        else
            return false;
    }

    static class pair implements Comparable {
        int f;
        int s;

        pair(int fi, int se) {
            f = fi;
            s = se;
        }

        public int compareTo(Object o)//desc order
        {
            pair pr = (pair) o;
            if (s > pr.s)
                return -1;
            if (s == pr.s) {
                if (f > pr.f)
                    return 1;
                else
                    return -1;
            } else
                return 1;
        }

        public boolean equals(Object o) {
            pair ob = (pair) o;
            if (o != null) {
                if ((ob.f == this.f) && (ob.s == this.s))
                    return true;
            }
            return false;
        }

        public int hashCode() {
            return (this.f + " " + this.s).hashCode();
        }


    }

    public static boolean palin(int l, int r, char[] c) {
        while (l <= r) {
            if (c[l] != c[r]) return false;
            l++;
            r--;
        }
        return true;
    }


    public static long gcd(long a, long b) {
        if (b == 0)
            return a;

        return gcd(b, a % b);
    }
    public static long lcm(long a, long b)
    {
        return (a*b)/gcd(a, b);
    }

    public static long hcf(long a, long b) {
        long t;
        while (b != 0) {
            t = b;
            b = a % b;
            a = t;
        }
        return a;
    }

    public static boolean isPrime(long n) {
        if (n <= 1)
            return false;

        // Check from 2 to n-1
        for (int i = 2; i <= Math.sqrt(n) + 1; i++)
            if (n % i == 0)
                return false;

        return true;
    }

    public static String reverse(String str) {
        String str1 = "";
        for (int i = 0; i < str.length(); i++) {
            str1 = str1 + str.charAt(str.length() - i - 1);
        }
        return str1;
    }

    public static double fact(long a) {
        if (a == 1)
            return 1;
        else
            return a * fact(a - 1);
    }

    static boolean isPerfectSquare(double x)
    {

        // Find floating point value of
        // square root of x.
        double sr = Math.sqrt(x);

        // If square root is an integer
        return ((sr - Math.floor(sr)) == 0);
    }


    public static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }


    }


    public static void main(String[] args) {
        FastReader s = new FastReader();
        //System.out.println("Forces Babyy!!");


      int test = s.nextInt();
      while(test-->0) {

          int n=s.nextInt();
          int[] a=new int[n];
          for(int i=0;i<n;i++)
              a[i]=s.nextInt();
          int[][] aa=new int[n][];
          for(int i=0;i<n;i++)
          {
              aa[i] = new int[]{a[i],i};
          }
          Arrays.sort(aa, new Comparator<int[]>() {
              @Override
              public int compare(int[] ints, int[] t1) {
                  if(ints[0]!=t1[0])return ints[0]-t1[0];
                  return ints[1]-t1[1];
              }
          });
          /*for(int i=0;i<n;i++)
              System.out.println(aa[i][0] +" " + aa[i][1]);*/

          int ans=0;
          int temp=0;
          for(int i=0;i<n;i++)
          {
              if(i==0 || aa[i][1]>aa[i-1][1])
                  temp++;
              else
                  temp=1;
              ans=Math.max(ans,temp);
          }
          System.out.println(n-ans);

      }
    }


    public static void fun(int[] a, int n, int x)
    {
        int temp = a[x];
        for(int i=x;i>=1;i--)
        {
            a[i]=a[i-1];
        }
        a[0]=temp;
    }

    public static void fun1(int[] a, int n, int x)
    {
        int temp=a[x];
        for(int i=x;i<n-1;i++)
        {
            a[i]=a[i+1];
        }
        a[n-1]=temp;
    }


}