Problem: There is a road with length l meters. The start of the road has coordinate 0, the end of the road has coordinate l.

There are two cars, the first standing at the start of the road and the second standing at the end of the road. They will start driving simultaneously. The first car will drive from the start to the end and the second car will drive from the end to the start.

Initially, they will drive with a speed of 1 meter per second. There are n flags at different coordinates a_1, a_2, …, a_n. Each time when any of two cars drives through a flag, the speed of that car increases by 1 meter per second.

Find how long will it take for cars to meet (to reach the same coordinate). 

Input

The first line contains one integer t (1 ≤ t ≤ 10^4): the number of test cases.

The first line of each test case contains two integers n, l (1 ≤ n ≤ 10^5, 1 ≤ l ≤ 10^9): the number of flags and the length of the road.

The second line contains n integers a_1, a_2, …, a_n in the increasing order (1 ≤ a_1 < a_2 < … < a_n < l).

It is guaranteed that the sum of n among all test cases does not exceed 10^5.

Output

For each test case print a single real number: the time required for cars to meet.

Your answer will be considered correct, if its absolute or relative error does not exceed 10^{-6}. More formally, if your answer is a and jury's answer is b, your answer will be considered correct if \frac{|a-b|}{max{(1, b)}} ≤ 10^{-6}.

Example

Input


5
2 10
1 9
1 10
1
5 7
1 2 3 4 6
2 1000000000
413470354 982876160
9 478
1 10 25 33 239 445 453 468 477


Output


3.000000000000000
3.666666666666667
2.047619047619048
329737645.750000000000000
53.700000000000000

Note

In the first test case cars will meet in the coordinate 5.

The first car will be in the coordinate 1 in 1 second and after that its speed will increase by 1 and will be equal to 2 meters per second. After 2 more seconds it will be in the coordinate 5. So, it will be in the coordinate 5 in 3 seconds.

The second car will be in the coordinate 9 in 1 second and after that its speed will increase by 1 and will be equal to 2 meters per second. After 2 more seconds it will be in the coordinate 5. So, it will be in the coordinate 5 in 3 seconds.

In the second test case after 1 second the first car will be in the coordinate 1 and will have the speed equal to 2 meters per second, the second car will be in the coordinate 9 and will have the speed equal to 1 meter per second. So, they will meet after (9-1)/(2+1) = 8/3 seconds. So, the answer is equal to 1 + 8/3 = 11/3.

Solution (in JAVA):

Solution:
import java.util.*;
import java.lang.*;
import java.io.*;  
public class Main 
{
/*
    HashMap<> map=new HashMap<>();
    TreeMap<> map=new TreeMap<>();
    map.put(p,map.getOrDefault(p,0)+1);
    for(Map.Entry<> mx:map.entrySet()){
          int v=mx.getValue(),k=mx.getKey();
    }for (int i = 1; i <= 1000; i++)
     fac[i] = fac[i - 1] * i % mod;
    ArrayList<Pair<Character,Integer>> l=new ArrayList<>();
    ArrayList<> l=new ArrayList<>();
    HashSet<> has=new HashSet<>();*/ 
    PrintWriter out;
    FastReader sc;
    long mod=(long)(1e9+7);
    int maxint= Integer.MAX_VALUE;
    int minint= Integer.MIN_VALUE;
    long maxlong=Long.MAX_VALUE;
    long minlong=Long.MIN_VALUE;
    public void sol(){
          int testCase=ni();  
          while(testCase-->0){
                int n=ni(),l=ni();
                int[] ar=new int[n+2];
                ar[0]=0;
                for(int i=1;i<=n;i++){
                      ar[i]=ni();
                }ar[n+1]=l;
                double[][] a=new double[n+2][2],b=new double[n+2][2];
                a[0][1]=1;b[n+1][1]=1;a[0][0]=0;b[n+1][0]=0;
                int s1=1,d1=0,s2=1,d2=l;double t1=0,t2=0;
                for(int i=1;i<=n+1;i++){
                      int p=ar[i];
                      int q=ar[n-i+1];
                      a[i][0]=t1+(((p-d1)*1d)/(1d*s1));
                      b[n-i+1][0]=t2+(((d2-q)*1d)/(1d*s2));
                      a[i][1]=s1+1;
                      b[n-i+1][1]=s2+1;
                      s1++;
                      d1=p;
                      t1=a[i][0];
                      s2++;
                      d2=q;
                      t2=b[n-i+1][0];
                }double ans=0;
            //     for(int i=0;i<=n+1;i++)pl(a[i][0]+" "+b[i][0]);
                for(int i=1;i<=n+1;i++){
                      if(a[i][0]>b[i][0]){
                         t1=a[i-1][0];t2=b[i][0];
                         double x1=ar[i-1],x2=ar[i];
                         if(t1>t2){
                              x2-=((t1-t2)*1d*b[i][1]); 
                         }else{
                               x1+=((t2-t1)*1d*(a[i-1][1]));
                         }t1=max(t1,t2);
                         double sp=b[i][1]+a[i-1][1];
                         t1+=((x2-x1)/sp);
                         ans=t1;
                         break;
                      }else if(a[i][0]==b[i][0]){
                            ans=a[i][0];
                            break;
                      }
                }pl(ans);
          }
    }
    public static void main(String[] args) 
	{ 
	      Main g=new Main();
		g.out=new PrintWriter(System.out);
		g.sc=new FastReader();
		g.sol();
		g.out.flush();
	}
	static class FastReader 
	{ 
		BufferedReader br; 
		StringTokenizer st; 
 
		public FastReader() 
		{ 
			br = new BufferedReader(new
					InputStreamReader(System.in)); 
		} 
 
		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()); 
		} 
 
		String nextLine() 
		{ 
			String str = ""; 
			try
			{ 
				str = br.readLine(); 
			} 
			catch (IOException e) 
			{ 
				e.printStackTrace(); 
			} 
			return str; 
		} 
	} public int ni(){
	      return sc.nextInt();
	}public long nl(){
	      return sc.nextLong();
	}public double nd(){
	      return sc.nextDouble();
	}public char[] rl(){
	      return sc.nextLine().toCharArray();
	}public String rl1(){
	      return sc.nextLine();
	}
	public void pl(Object s){
	      out.println(s);
	}public void ex(){
	      out.println();
	}
	public void pr(Object s){
	      out.print(s);
	}public String next(){
	      return sc.next();
	}public long abs(long x){
	      return Math.abs(x);
	}
	public int abs(int x){
	      return Math.abs(x);
	}
	public double abs(double x){
	      return Math.abs(x);
	}
      public long pow(long x,long y){
            return (long)Math.pow(x,y);
      }
      public int pow(int x,int y){
            return (int)Math.pow(x,y);
      }
      public double pow(double x,double y){
            return Math.pow(x,y);
      }public long min(long x,long y){
            return (long)Math.min(x,y);
      }
      public int min(int x,int y){
            return (int)Math.min(x,y);
      }
      public double min(double x,double y){
            return Math.min(x,y);
      }public static int gcd(int a, int b) {
		if (a == 0)
			return b;
		return gcd(b % a, a);
	}static void sort(int[] a) {
		ArrayList<Integer> l = new ArrayList<>();
		for (int i : a) {
			l.add(i);
		}
		Collections.sort(l);
		for (int i = 0; i < a.length; i++) {
			a[i] = l.get(i);
		}
	}void sort(long[] a) {
		ArrayList<Long> l = new ArrayList<>();
		for (long i : a) {
			l.add(i);
		}
		Collections.sort(l);
		for (int i = 0; i < a.length; i++) {
			a[i] = l.get(i);
		}
	}void sort(double[] a) {
		ArrayList<Double> l = new ArrayList<>();
		for (double i : a) {
			l.add(i);
		}
		Collections.sort(l);
		for (int i = 0; i < a.length; i++) {
			a[i] = l.get(i);
		}
	}int  swap(int a,int b){
	      return a;
	}long  swap(long a,long b){
	      return a;
	}double  swap(double a,double b){
	      return a;
	}
	boolean isPowerOfTwo (int x) 
      { 
            return x!=0 && ((x&(x-1)) == 0);     
      }boolean isPowerOfTwo (long x) 
      { 
            return x!=0 && ((x&(x-1)) == 0);     
      }public long max(long x,long y){
            return (long)Math.max(x,y);
      }
      public int max(int x,int y){
            return (int)Math.max(x,y);
      }
      public double max(double x,double y){
            return Math.max(x,y);
      }long sqrt(long x){
            return (long)Math.sqrt(x);
      }int sqrt(int x){
            return (int)Math.sqrt(x);
      }void input(int[] ar,int n){
            for(int i=0;i<n;i++)ar[i]=ni();
      }void input(long[] ar,int n){
            for(int i=0;i<n;i++)ar[i]=nl();
      }void fill(int[] ar,int k){
            Arrays.fill(ar,k);
      }void yes(){
            pl("YES");
      }void no(){
            pl("NO");
      }
    int[] sieve(int n) 
    { 
        boolean prime[] = new boolean[n+1]; 
        int[] k=new int[n+1];
        for(int i=0;i<=n;i++) {
            prime[i] = true; 
            k[i]=i;
        }
          
        for(int p = 2; p <=n; p++) 
        { 
            if(prime[p] == true) 
            { 
               // sieve[p]=p;
                for(int i = p*2; i <= n; i += p) {
                    prime[i] = false; 
                  //  sieve[i]=p;
                    while(k[i]%(p*p)==0){
                          k[i]/=(p*p);
                    }
                }
            } 
        }return k;
    }int strSmall(int[] arr, int target)  
    {  
        int start = 0, end = arr.length-1;  
        int ans = -1;  
        while (start <= end) {  
            int mid = (start + end) / 2;   
            if (arr[mid] >= target) {  
                end = mid - 1;  
            }   
            else {  
                ans = mid;  
                start = mid + 1;  
            }  
        }  
        return ans;  
    } int strSmall(ArrayList<Integer> arr, int target)  
    {  
        int start = 0, end = arr.size()-1;  
        int ans = -1;  
        while (start <= end) {  
            int mid = (start + end) / 2;   
            if (arr.get(mid) > target) {  
                start = mid + 1; 
                ans=start;
            }   
            else {
                end = mid - 1;  
            }  
        }  
        return ans;  
    }long mMultiplication(long a,long b) 
    { 
        long res = 0;    
        a %= mod; 
  
        while (b > 0)  
        {  
            if ((b & 1) > 0)  
            { 
                res = (res + a) % mod; 
            }  
            a = (2 * a) % mod; 
  
            b >>= 1;  
        } 
        return res; 
    }long nCr(int n, int r,
                             long p)
    {
 
          if (n<r) 
              return 0;
        if (r == 0)
            return 1;
            long[] fac = new long[n + 1];
        fac[0] = 1;
 
        for (int i = 1; i <= n; i++)
            fac[i] = fac[i - 1] * i % p;
        return (fac[n] * modInverse(fac[r], p)
                % p * modInverse(fac[n - r], p)
                % p)
            % p;
    }long power(long x, long y, long p)
    {
        long res = 1;
        x = x % p;
 
        while (y > 0) {
            if (y % 2 == 1)
                res = (res * x) % p;
            y = y >> 1; 
            x = (x * x) % p;
        }
 
        return res;
    }long modInverse(long n, long p)
    {
        return power(n, p - 2, p);
    }
      public static class Pair implements Comparable<Pair> {
		int x;
		int y;
 
		public Pair(int x, int y) {
			this.x = x;
			this.y = y;
		}
 
		public String toString() {
			return x + "," + y;
		}
 
		public boolean equals(Object o) {
			if (o instanceof Pair) {
				Pair p = (Pair) o;
				return p.x == x && p.y == y;
			}
			return false;
		}
 
		public int hashCode() {
			return new Double(x).hashCode() * 31 + new Double(y).hashCode();
		}
 
		public int compareTo(Pair other) {
			if (this.x == other.x) {
				return Long.compare(this.y, other.y);
			}
			return Long.compare(this.x, other.x);
		}
      }
	 
}