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.*;
public class Main {
    static class Scan {
        private byte[] buf=new byte[1024];
        private int index;
        private InputStream in;
        private int total;
        public Scan()
        {
            in=System.in;
        }
        public int scan()throws IOException
        {
            if(total<0)
            throw new InputMismatchException();
            if(index>=total)
            {
                index=0;
                total=in.read(buf);
                if(total<=0)
                return -1;
            }
            return buf[index++];
        }
        public int scanInt()throws IOException
        {
            int integer=0;
            int n=scan();
            while(isWhiteSpace(n))
            n=scan();
            int neg=1;
            if(n=='-')
            {
                neg=-1;
                n=scan();
            }
            while(!isWhiteSpace(n))
            {
                if(n>='0'&&n<='9')
                {
                    integer*=10;
                    integer+=n-'0';
                    n=scan();
                }
                else throw new InputMismatchException();
            }
            return neg*integer;
        }
        public double scanDouble()throws IOException
        {
            double doub=0;
            int n=scan();
            while(isWhiteSpace(n))
            n=scan();
            int neg=1;
            if(n=='-')
            {
                neg=-1;
                n=scan();
            }
            while(!isWhiteSpace(n)&&n!='.')
            {
                if(n>='0'&&n<='9')
                {
                    doub*=10;
                    doub+=n-'0';
                    n=scan();
                }
                else throw new InputMismatchException();
            }
            if(n=='.')
            {
                n=scan();
                double temp=1;
                while(!isWhiteSpace(n))
                {
                    if(n>='0'&&n<='9')
                    {
                        temp/=10;
                        doub+=(n-'0')*temp;
                        n=scan();
                    }
                    else throw new InputMismatchException();
                }
            }
            return doub*neg;
        }
        public String scanString()throws IOException
        {
            StringBuilder sb=new StringBuilder();
            int n=scan();
            while(isWhiteSpace(n))
            n=scan();
            while(!isWhiteSpace(n))
            {
                sb.append((char)n);
                n=scan();
            }
            return sb.toString();
        }
        private boolean isWhiteSpace(int n)
        {
            if(n==' '||n=='\n'||n=='\r'||n=='\t'||n==-1)
            return true;
            return false;
        }
    }
    
    public static void sort(int arr[],int l,int r) {    //sort(arr,0,n-1);
        if(l==r) {
            return;
        }
        int mid=(l+r)/2;
        sort(arr,l,mid);
        sort(arr,mid+1,r);
        merge(arr,l,mid,mid+1,r);
    }
    public static void merge(int arr[],int l1,int r1,int l2,int r2) {
        int tmp[]=new int[r2-l1+1];
        int indx1=l1,indx2=l2;
        //sorting the two halves using a tmp array
        for(int i=0;i<tmp.length;i++) {
            if(indx1>r1) {
                tmp[i]=arr[indx2];
                indx2++;
                continue;
            }
            if(indx2>r2) {
                tmp[i]=arr[indx1];
                indx1++;
                continue;
            }
            if(arr[indx1]<arr[indx2]) {
                tmp[i]=arr[indx1];
                indx1++;
                continue;
            }
            tmp[i]=arr[indx2];
            indx2++;
        }
        //Copying the elements of tmp into the main array
        for(int i=0,j=l1;i<tmp.length;i++,j++) {
            arr[j]=tmp[i];
        }
    }
    
    public static void sort(long arr[],int l,int r) {    //sort(arr,0,n-1);
        if(l==r) {
            return;
        }
        int mid=(l+r)/2;
        sort(arr,l,mid);
        sort(arr,mid+1,r);
        merge(arr,l,mid,mid+1,r);
    }
    public static void merge(long arr[],int l1,int r1,int l2,int r2) {
        long tmp[]=new long[r2-l1+1];
        int indx1=l1,indx2=l2;
        //sorting the two halves using a tmp array
        for(int i=0;i<tmp.length;i++) {
            if(indx1>r1) {
                tmp[i]=arr[indx2];
                indx2++;
                continue;
            }
            if(indx2>r2) {
                tmp[i]=arr[indx1];
                indx1++;
                continue;
            }
            if(arr[indx1]<arr[indx2]) {
                tmp[i]=arr[indx1];
                indx1++;
                continue;
            }
            tmp[i]=arr[indx2];
            indx2++;
        }
        //Copying the elements of tmp into the main array
        for(int i=0,j=l1;i<tmp.length;i++,j++) {
            arr[j]=tmp[i];
        }
    }
    static int n,arr[],brr[],dp[][][];
    static TreeMap<Integer,TreeSet<Integer>> tm;
    public static void main(String args[]) throws IOException {
        Scan input=new Scan();
        StringBuilder ans=new StringBuilder("");
        int test=input.scanInt();
        for(int tt=1;tt<=test;tt++) {
            n=input.scanInt();
            arr=new int[n];
            brr=new int[n];
            for(int i=0;i<n;i++) {
                arr[i]=input.scanInt();
                brr[i]=arr[i];
            }
            sort(brr,0,n-1);
            HashMap<Integer,Integer> map=new HashMap<>();
            int cnt=0;
            for(int i=0;i<n;i++) {
                if(!map.containsKey(brr[i])) {
                    map.put(brr[i], cnt);
                    cnt++;
                }
            }
            for(int i=0;i<n;i++) {
                arr[i]=map.get(arr[i]);
            }
//            for(int i=0;i<n;i++) {
//                System.out.print(arr[i]+" ");
//            }
//            System.out.println();
            dp=new int[n][2][2];
            for(int i=0;i<n;i++) {
                for(int j=0;j<2;j++) {
                    for(int k=0;k<2;k++) {
                        dp[i][j][k]=-1;
                    }
                }
            }
            tm=new TreeMap<>();
            for(int i=0;i<n;i++) {
                if(!tm.containsKey(arr[i])) {
                    TreeSet<Integer> tmp=new TreeSet<>();
                    tm.put(arr[i], tmp);
                }
                TreeSet<Integer> tmp=tm.get(arr[i]);
                tmp.add(i);
            }
            int max=0;
            for(int i=0;i<n;i++) {
                max=Math.max(max,solve(i,true,false));
//                System.out.println(i+" "+solve(i,true,false));
            }
            ans.append(n-max);
            ans.append("\n");
        }
        System.out.println(ans);
    }
    public static int solve(int indx,boolean strt,boolean end) {
        if(dp[indx][strt?1:0][end?1:0]!=-1) {
            return dp[indx][strt?1:0][end?1:0];
        } 
        int max=1;
        if(strt) {
            TreeSet<Integer> tmp=tm.get(arr[indx]);
            if(tmp.higher(indx)!=null) {
                max=Math.max(max,1+solve(tmp.higher(indx),strt,end));
            }
            if(tm.containsKey(arr[indx]+1)) {
                tmp=tm.get(arr[indx]+1);
                if(tmp.higher(indx)!=null && tmp.first()==tmp.higher(indx)) {
                    max=Math.max(max,tmp.size()+solve(tmp.last(),false,false));
                }
                if(tmp.higher(indx)!=null && tmp.first()!=tmp.higher(indx)) {
                    max=Math.max(max,1+solve(tmp.higher(indx),false,true));
                }
            }
        }
        if(!strt && !end) {
            TreeSet<Integer> tmp;
            if(tm.containsKey(arr[indx]+1)) {
                tmp=tm.get(arr[indx]+1);
                if(tmp.higher(indx)!=null && tmp.first()==tmp.higher(indx)) {
                    max=Math.max(max,tmp.size()+solve(tmp.last(),false,false));
                }
                if(tmp.higher(indx)!=null && tmp.first()!=tmp.higher(indx)) {
                    max=Math.max(max,1+solve(tmp.higher(indx),false,true));
                }
            }
        }
        if(end) {
            TreeSet<Integer> tmp=tm.get(arr[indx]);
            if(tmp.higher(indx)!=null) {
                max=Math.max(max,1+solve(tmp.higher(indx),false,true));
            }
        }
        dp[indx][strt?1:0][end?1:0]=max;
        return max;
    }
}