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.io.*;
import java.util.*;
import java.lang.*;
import java.math.*;
import java.text.DecimalFormat;
import java.lang.reflect.Array;
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigDecimal;
public class B{
	public static PrintWriter out = new PrintWriter (new BufferedOutputStream(System.out));
	//static long MOD = (long)(1e9+7);
	static int MOD = 998244353;
	static long MOD2 = MOD*MOD;
	static FastReader sc = new FastReader();
	static int pInf = Integer.MAX_VALUE;
	static int nInf = Integer.MIN_VALUE;
	//static ArrayList<Long> A;
	public static void main(String[] args){
		int test = 1;
		test = sc.nextInt();
		while(test-->0)solve();
		out.flush();
		out.close();
	}
	//static ArrayList<Integer>[] A;
	public static void solve() {
		int n = sc.nextInt();
		int l = sc.nextInt();
		double[] pos = new double[n];
		for(int i = 0; i < n; i++) {
			pos[i] = sc.nextDouble();
		}
		double lPos = 0,rPos = l,lSpeed = 1,rSpeed = 1,totTime = 0;
		int lPointer = 0,rPointer = n-1;
		while(lPointer<=rPointer) {
			double ldist = pos[lPointer]-lPos,rdist = rPos-pos[rPointer];
			double ltime = ldist/lSpeed,rtime = rdist/rSpeed;
			if(ltime<rtime) {
				lPos = pos[lPointer];
				lSpeed+=1;
				rPos -= rSpeed*ltime;
				totTime += ltime;	
				lPointer++;
			}
			else {
				rPos = pos[rPointer];
				rSpeed+=1;
				lPos += lSpeed*rtime;
				totTime += rtime;
				rPointer--;
			}
		}
		totTime+= (rPos-lPos)/(rSpeed+lSpeed);
		out.printf("%.10f\n", totTime);
		return;
	}
	public static int maxRight(int x,int[] a) {
		int l = -1;      //a[l]<=x
		int r = a.length;//a[r]>x
		while(r-l>1) {
			int m = l+(r-l)/2;
			if(a[m]<=x) {
				l = m;
			}
			else {
				r = m;
			}
		}
		return l+1;
	}
	public static int minLeft(int x,int[] a) {
		int l = -1;      //a[l]<x
		int r = a.length;//a[r]>=x
		while(r-l>1) {
			int m = l+(r-l)/2;
			if(a[m]<x) {
				l = m;
			}
			else {
				r = m;
			}
		}
		return r+1;
	}
	public static int lowerBound(int key, int[] a) {
		int s = 0;
		int e = a.length-1;
		if(e==-1) {
			return 0;
		}
		int ans = -1;
		while(s<=e) {
			int m = s+(e-s)/2;
			if(a[m]>=key) {
				ans = m;
				e = m-1;
			}
			else {
				s = m+1;
			}
		}
		return ans==-1?s:ans;
	}
	public static int upperBound(int key, int[] a) {
		int s = 0;
		int e = a.length-1;
		if(e==-1) {
			return 0;
		}
		int ans = -1;
		while(s<=e) {
			int m = s+(e-s)/2;
			if(a[m]>key) {
				ans = m;
				e = m-1;
			}
			else {
				s = m+1;
			}
		}
		return ans==-1?s:ans;
	}
	public static long c2(long n) {
		if((n&1)==0) {
			return mul(n/2, n-1);
		}
		else {
			return mul(n, (n-1)/2);
		}
	}
	public static long mul(long a, long b){
	    return ((a%MOD)*(b%MOD))%MOD;
	}
	public static long add(long a, long b){
	    return ((a%MOD)+(b%MOD))%MOD;
	}
	public static long sub(long a, long b){
	    return ((a%MOD)-(b%MOD))%MOD;
	}
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
//Integer.lowestOneBit(i)   Equals k where k is the position of the first one in the binary
//Integer.highestOneBit(i)  Equals k where k is the position of the last one in the binary
//Integer.bitCount(i)       returns the number of one-bits
//Collections.sort(A,(p1,p2)->(int)(p2.x-p1.x)) To sort ArrayList in descending order wrt values of x.
//            Arrays.parallelSort(a,new Comparator<TPair>() {
//      		public int compare(TPair a,TPair b) {
//        			if(a.y==b.y) return a.x-b.x;
//        			return b.y-a.y;
//        		}
//        	});
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
//PrimeFactors    
    public static ArrayList<Long> primeFactors(long n) {
        ArrayList<Long> arr = new ArrayList<>();
        if (n % 2 == 0)
            arr.add((long) 2);
        while (n % 2 == 0)
            n /= 2;
        for (long i = 3; i <= Math.sqrt(n); i += 2) {
            int flag = 0;
            while (n % i == 0) {
                n /= i;
                flag = 1;
            }
            if (flag == 1)
                arr.add(i);
        }
        if (n > 2)
            arr.add(n);
        return arr;
    }
//Pair Class
	static class Pair implements Comparable<Pair>{
		int x;
		int y;
		public Pair(int x, int y) {
			this.x = x;
			this.y = y;
		}
		@Override
		public int compareTo(Pair o) {
			if(this.x==o.x){
			    return (this.y-o.y);
			}
			return (this.x-o.x);
		}
	}
	static class TPair implements Comparable<TPair>{
		int l;
		int r;
		int index;
		public TPair(int l, int r, int index) {
			this.l = l;
			this.r = r;
			this.index = index;
		}
		@Override
		public int compareTo(TPair o) {
			if(this.l==o.l){
			    return -1*(this.r-o.r);
			}
			return -1*(this.l-o.l);
		}
	}
//nCr
	static long[] facts, factInvs;
	static long exp(long base, long e) {
		if (e==0) return 1;
		long half=exp(base, e/2);
		if (e%2==0) return mul(half, half);
		return mul(half, mul(half, base));
	}
	
	static long modInv(long x) {
		return exp(x, MOD-2);
	}
	
	static void precomp() {
		facts=new long[1_000_000];
		factInvs=new long[1_000_000];
		factInvs[0]=facts[0]=1;
		for (int i=1; i<facts.length; i++)
			facts[i]=mul(facts[i-1], i);
		factInvs[facts.length-1]=modInv(facts[facts.length-1]);
		for (int i=facts.length-2; i>=0; i--)
			factInvs[i]=mul(factInvs[i+1], i+1);
	}
	
	static long nCk(int n, int k) {
		return mul(facts[n], mul(factInvs[k], factInvs[n-k]));
	}
//Kadane's Algorithm    
    static long maxSubArraySum(ArrayList<Long> a) { 
    	if(a.size()==0) {
    		return 0;
    	}
    	long max_so_far = a.get(0); 
    	long curr_max = a.get(0); 
    	for (int i = 1; i < a.size(); i++) { 
           curr_max = Math.max(a.get(i), curr_max+a.get(i)); 
           max_so_far = Math.max(max_so_far, curr_max); 
    	} 
    	return max_so_far; 
    } 
//Shuffle Sort
    static final Random random=new Random();
	static void ruffleSort(int[] a) {
		int n=a.length;//shuffle, then sort 
		for (int i=0; i<n; i++) {
			int oi=random.nextInt(n), temp=a[oi];
			a[oi]=a[i]; a[i]=temp;
		}
		Arrays.sort(a);
	}
//Merge Sort
	static void merge(long arr[], int l, int m, int r)
    {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;
 
        /* Create temp arrays */
        long L[] = new long [n1];
        long R[] = new long [n2];
 
        /*Copy data to temp arrays*/
        for (int i=0; i<n1; ++i)
            L[i] = arr[l + i];
        for (int j=0; j<n2; ++j)
            R[j] = arr[m + 1+ j];
 
 
        /* Merge the temp arrays */
 
        // Initial indexes of first and second subarrays
        int i = 0, j = 0;
 
        // Initial index of merged subarry array
        int k = l;
        while (i < n1 && j < n2)
        {
            if (L[i] <= R[j])
            {
                arr[k] = L[i];
                i++;
            }
            else
            {
                arr[k] = R[j];
                j++;
            }
            k++;
        }
 
        /* Copy remaining elements of L[] if any */
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }
 
        /* Copy remaining elements of R[] if any */
        while (j < n2)
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }
 
    // Main function that sorts arr[l..r] using
    // merge()
    static void sort(long arr[], int l, int r)
    {
        if (l < r)
        {
            // Find the middle point
            int m = (l+r)/2;
 
            // Sort first and second halves
            sort(arr, l, m);
            sort(arr , m+1, r);
 
            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }
//Brian Kernighans Algorithm
    static long countSetBits(long n){
        if(n==0) return 0;
        return 1+countSetBits(n&(n-1));
    }
//Factorial
    static long factorial(long n){
        if(n==1) return 1;
        if(n==2) return 2;
        if(n==3) return 6;
        return n*factorial(n-1);
    }
//Euclidean Algorithm
    static long gcd(long A,long B){
        if(B==0) return A;
        return gcd(B,A%B);
    }
//Modular Exponentiation
    static long fastExpo(long x,long n){
        if(n==0) return 1;
        if((n&1)==0) return fastExpo((x*x)%MOD,n/2)%MOD;
        return ((x%MOD)*fastExpo((x*x)%MOD,(n-1)/2))%MOD;
    }
//AKS Algorithm
    static boolean isPrime(long n){
        if(n<=1) return false;
        if(n<=3) return true;
        if(n%2==0 || n%3==0) return false;
        for(int i=5;i<=Math.sqrt(n);i+=6)
            if(n%i==0 || n%(i+2)==0) return false;
        return true;
    }
//Reverse an array
    static <T> void reverse(T arr[],int l,int r){
    	Collections.reverse(Arrays.asList(arr).subList(l, r));
    }
  //Print array
    static void print1d(long arr[]) {
    	out.println(Arrays.toString(arr));
    }
    static void print2d(int arr[][]) {
    	for(int a[]: arr) out.println(Arrays.toString(a));
    }
//Sieve of eratosthenes
    static int[] findPrimes(int n){
        boolean isPrime[]=new boolean[n+1];
        ArrayList<Integer> a=new ArrayList<>();
        int result[];
        Arrays.fill(isPrime,true);
        isPrime[0]=false;
        isPrime[1]=false;
        for(int i=2;i*i<=n;++i){
            if(isPrime[i]==true){
                for(int j=i*i;j<=n;j+=i) isPrime[j]=false;
            }
        }
        for(int i=0;i<=n;i++) if(isPrime[i]==true) a.add(i);
        result=new int[a.size()];
        for(int i=0;i<a.size();i++) result[i]=a.get(i);
        return result;
        
    }
//Indivisual factors of all nos till n
    static ArrayList<Integer>[] indiFactors(int n){
    	ArrayList<Integer>[] A = new ArrayList[n+1];
    	for(int i = 0; i <= n; i++) {
    		A[i] = new ArrayList<>();
    	}
    	int[] sieve = new int[n+1];
    	for(int i=2;i<=n;i++) {
       		if(sieve[i]==0) {
       			for(int j=i;j<=n;j+=i) if(sieve[j]==0) {
       				//sieve[j]=i;
       				A[j].add(i);
       			}
       		}
       	}
    	return A;
    }
//Segmented Sieve
    static boolean[] segmentedSieve(long l, long r){
    	boolean[] segSieve = new boolean[(int)(r-l+1)];
    	Arrays.fill(segSieve, true);
    	int[] prePrimes = findPrimes((int)Math.sqrt(r));
    	for(int p:prePrimes) {
    		long low = (l/p)*p;
    		if(low < l) {
    			low += p;
    		}
    		if(low == p) {
    			low += p;
    		}
    		for(long j = low; j<= r; j += p) {
    			segSieve[(int) (j-l)] = false;
    		}
    	}
    	if(l==1) {
    		segSieve[0] = false;
    	}
    	return segSieve;
    }
//Euler Totent function
    static long countCoprimes(long n){
        ArrayList<Long> prime_factors=new ArrayList<>();
        long x=n,flag=0;
        while(x%2==0){
            if(flag==0) prime_factors.add(2L);
            flag=1;
            x/=2;
        }
        for(long i=3;i*i<=x;i+=2){
            flag=0;
            while(x%i==0){
                if(flag==0) prime_factors.add(i);
                flag=1;
                x/=i;
            }
        }
        if(x>2) prime_factors.add(x);
        double ans=(double)n;
        for(Long p:prime_factors){
            ans*=(1.0-(Double)1.0/p);
        }
        return (long)ans;
    }
	static long modulo = (long)1e9+7;
	public static long modinv(long x){
	    return modpow(x, modulo-2);
	}
	public static long modpow(long a, long b){
	    if(b==0){
	        return 1;
	    }
	    long x = modpow(a, b/2);
	    x = (x*x)%modulo;
	    if(b%2==1){
	        return (x*a)%modulo;
	    }
	    return x;
	}
	public static class FastReader {
		BufferedReader br;
		StringTokenizer st;
		//it reads the data about the specified point and divide the data about it ,it is quite fast
		//than using direct 
 
		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
 
		String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (Exception r) {
					r.printStackTrace();
				}
			}
			return st.nextToken();
		}
 
		int nextInt() {
			return Integer.parseInt(next());//converts string to integer
		}
 
		double nextDouble() {
			return Double.parseDouble(next());
		}
 
		long nextLong() {
			return Long.parseLong(next());
		}
 
		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (Exception r) {
				r.printStackTrace();
			}
			return str;
		}
	}
}