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.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.Random;
import java.util.StringTokenizer;
import java.util.TreeSet;
 
public class Main {

static class Pair implements Comparable<Pair>{
	int value, index;
	public Pair(int value, int index) {
		this.value = value;
		this.index = index;
	}
	@Override
	public int compareTo(Pair o) {
		return Integer.compare(value, o.value);
	}
}
	
public static void main(String[] args){
 
   new Thread(null, null, "Anshum Gupta", 99999999) {
        public void run() {
            try {
                solve();
            } catch(Exception e) {
                e.printStackTrace();
                System.exit(1);
            }
        }
    }.start();
}
 
static final long mxx = (long)(1e18 + 5);
static final int mxN = (int)(1e6);
static final int mxV = (int)(1e6), log = 18;
static final long MOD = 998244353; 
static final int INF = (int)1e9;
static boolean[]vis;
static ArrayList<ArrayList<Integer>> adj;
static int n, m, q, k;
static char[]str;

public static void solve() throws Exception {
	   // solve the problem here
		s = new MyScanner();
   		out = new PrintWriter(new BufferedOutputStream(System.out), true);
//	   		out = new PrintWriter("output.txt");
        int tc = s.nextInt();
        for(int i=1; i<=tc; i++) {
//        	out.print("Case #" + i + ": ");
        	testcase();
        }
           
        out.flush();
        out.close();
}




static void testcase() {
	n = s.nextInt();
	int[] a = s.nextIntArray(n);
	
	
	// now each element is number according to its position in the sorted array 
	// now each element in a is b/w [0, n-1]
	cordinateCompress(a); 
	
	int ans = 0;
	ArrayList<Integer>[] indices = new ArrayList[n];
	for(int i = 0; i < n; i++) indices[i] = new ArrayList<Integer>();
	for(int i = 0; i < n; i++) indices[a[i]].add(i);
	for(int i = 0; i < n; i++) ans = Math.max(ans, indices[i].size());
	
	int[] dp = new int[n];
	dp[n - 1] = indices[n - 1].size();
	for(int i = n - 2; i >= 0; i--) {
		if(indices[i].size() == 0)continue;
		dp[i] = indices[i].size();
		if(indices[i + 1].size() == 0)continue;
		if(indices[i + 1].get(0) > indices[i].get(indices[i].size() - 1))
			dp[i] += dp[i + 1];
		else
			dp[i] += howManyInRange(indices[i + 1], indices[i].get(indices[i].size() - 1), n - 1);
		
	}
	
	for(int i = 0; i < n; i++) {
		int prefix = i == 0 ? 0 : howManyInRange(indices[i - 1], 0, indices[i].isEmpty() ? -1 : indices[i].get(0) - 1);
		ans = Math.max(ans, dp[i] + prefix);
		if(a[i] < n - 1) {
			ans = Math.max(ans, howManyInRange(indices[a[i]], 0, i) + howManyInRange(indices[a[i] + 1], i + 1, n - 1));
		}
	}
	
	out.println(n - ans);
	
}


private static int howManyInRange(ArrayList<Integer> arr, int l, int r) {
	int len = arr.size();
	int lo = 0, hi = len - 1;
	int left = len;
	while(lo <= hi) {
		int mid = (lo + hi) >> 1;
		if(arr.get(mid) >= l) {
			left = mid;
			hi = mid - 1;
		}else lo = mid + 1;
	}
	int right = -1;
	lo = 0; hi = len - 1;
	while(lo <= hi) {
		int mid = (lo + hi) >> 1;
		if(arr.get(mid) <= r) {
			right = mid;
			lo = mid + 1;
		}else hi = mid - 1;
	}
	return Math.max(0, right - left + 1);
//	if(left == len || right == -1)return 0;
//	out.println("in range =>  [" + l + " , " + r + " ] => " + (right - left + 1));
//	return right - left + 1;
}


static int nInRange(ArrayList<Integer> list, int l, int r) {
//	System.out.println("Querying nInRange: "+l+" "+r+" for array: "+list);
	int n=list.size();
	if (list.isEmpty() || list.get(0)>r || list.get(n-1)<l || r<l) return 0;
	
	//first index >= l
	int min=0, max=n-1;
	while (min!=max) {
		int mid=(min+max)/2;
		if (list.get(mid) >= l) {
			max=mid;
		}
		else {
			min=mid+1;
		}
	}
	int firstLegal=min;
	
	//last index <=r
	min=0; max=n-1;
	while (min!=max) {
		int mid=(min+max+1)/2;
		if (list.get(mid)<=r) {
			min=mid;
		}
		else {
			max=mid-1;
		}
	}
	int lastLegal=min;
	return lastLegal-firstLegal+1;
}

private static void cordinateCompress(int[] a) {
	TreeSet<Integer> values = new TreeSet<>();
	for(int x : a)values.add(x);
	HashMap<Integer, Integer> mp = new HashMap<>();
	int index = 0;
	for(int x : values)mp.put(x, index++);
	for(int i = 0; i < n; i++)a[i] = mp.get(a[i]);
}

public static PrintWriter out;
public static MyScanner s;
static void shuffleArray(int[] a) {
	Random random = new Random();
	for (int i = a.length-1; i > 0; i--) {
		int index = random.nextInt(i + 1);
		int tmp = a[index];
		a[index] = a[i];
		a[i] = tmp;
	}
}
static void shuffleSort(int[] a) {
	shuffleArray(a);
	Arrays.parallelSort(a);
}
static class MyScanner {
 
    BufferedReader br;
    StringTokenizer st;
 
    public MyScanner() {
        br = new BufferedReader(new InputStreamReader(System.in));
    }
    public MyScanner(String fileName) {
    	try {
			br = new BufferedReader(new FileReader(fileName));
		} catch (FileNotFoundException e) {
			e.printStackTrace();
		}
    }
    String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }
 
    int nextInt() { return Integer.parseInt(next()); }
    long nextLong() { return Long.parseLong(next()); }
    double nextDouble() { return Double.parseDouble(next()); }
    int[] nextIntArray(int n){
    	int[]a = new int[n];
    	for(int i=0; i<n; i++) {
    		a[i] = this.nextInt();
    	}
    	return a;
    }
    long[] nextlongArray(int n) {
    	long[]a = new long[n];
    	for(int i=0; i<n; i++) {
    		a[i] = this.nextLong();
    	}
    	return a;
    }
    Integer[] nextIntegerArray(int n){
    	Integer[]a = new Integer[n];
    	for(int i=0; i<n; i++) {
    		a[i] = this.nextInt();
    	}
    	return a;
    }
    Long[] nextLongArray(int n) {
    	Long[]a = new Long[n];
    	for(int i=0; i<n; i++) {
    		a[i] = this.nextLong();
    	}
    	return a;
    }
    char[][] next2DCharArray(int n, int m){
    	char[][]arr = new char[n][m];
    	for(int i=0; i<n; i++) {
    		arr[i] = this.next().toCharArray();
    	}
    	return arr;
    }
    ArrayList<ArrayList<Integer>> readUndirectedUnweightedGraph(int n, int m) {
    	ArrayList<ArrayList<Integer>>adj = new ArrayList<ArrayList<Integer>>();
    	for(int i=0; i<=n; i++)adj.add(new ArrayList<Integer>());
    	for(int i=0; i<m; i++) {
    		int u = s.nextInt();
    		int v = s.nextInt();
    		adj.get(u).add(v);
    		adj.get(v).add(u);
    	}
    	return adj;
    }
    ArrayList<ArrayList<Integer>> readDirectedUnweightedGraph(int n, int m) {
    	ArrayList<ArrayList<Integer>>adj = new ArrayList<ArrayList<Integer>>();
    	for(int i=0; i<=n; i++)adj.add(new ArrayList<Integer>());
    	for(int i=0; i<m; i++) {
    		int u = s.nextInt();
    		int v = s.nextInt();
    		adj.get(u).add(v);
    	}
    	return adj;
    }
    String nextLine(){
        String str = "";
        try {
            str = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return str;
    }
}
 
 
}