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.math.BigDecimal;
import java.math.BigInteger;
import java.util.*;
 
public class Main {
	static final int INF = (int) (1e9 + 10);
	static final int MOD = (int) (1e9 + 7);
 
	// static final int N = (int) (4e5 + 5);
	// static ArrayList<Integer>[] graph;
	// static boolean visited[];
	// static long size[];
 
	public static void main(String[] args) throws NumberFormatException, IOException {
		FastReader sc = new FastReader();
		PrintWriter pr = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int t = sc.nextInt();
		while (t-- > 0) {
		    int n = sc.nextInt();
		    long l = sc.nextLong();
		    long [] a = new long [n];
		    
		    for(int i =0 ; i < n ; i++){
		        a[i] = sc.nextLong();
		      //  pr.print
		    }
		    double time =0;
		    int i =0;
		    int j = n-1;
		    double ld = 0;
		    double rd = (double)l;
		    int ls =1;
		    int rs =1;
		    while(i <= j){
		        double lt = Math.abs(ld-a[i])/ls;
		        double rt = Math.abs(rd-a[j])/rs;
		      //  ld+=a[i];
		      //  rd-=a[j];
		        
		        if(lt < rt){
		            rd-=(lt*rs);
		            ld +=(lt*ls);
		            ls++;
		            time += lt;
		            i++;
		        }else {
		            ld+=(rt*ls);
		            rd -=(rt*rs);
		            rs++;
		            time+=rt;
		            j--;
		        }
		    }
		    if(ld < rd)time+=Math.abs(rd-ld)/(rs+ls);
		    
		    
		    
		    pr.println(time);
		    
		    
		    
 
		}
		
		pr.flush();
		pr.close();
	}
 
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	private static boolean possible(long[] arr, int[] f, long mid, long k) {
		long c = 0;
 
		for (int i = 0; i < arr.length; i++) {
			if ((arr[i] * f[f.length - i - 1]) > mid) {
				c += (arr[i] - (mid / f[f.length - i - 1]));
			}
		}
		// System.out.println(mid+" "+c);
		if (c <= k)
			return true;
		return false;
	}
 
	public static int lowerBound(ArrayList<Integer> array, int length, long value) {
		int low = 0;
		int high = length;
		while (low < high) {
			final int mid = (low + high) / 2;
			if (value <= array.get(mid)) {
				high = mid;
			} else {
				low = mid + 1;
			}
		}
		return low;
	}
 
	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;
		}
	}
 
	static long gcd(long a, long b) {
		if (a == 0)
			return b;
		return gcd(b % a, a);
	}
 
	static long lcm(long a, long b) {
		return (a * b) / gcd(a, b);
	}
 
	public static void sortbyColumn(int[][] att, int col) {
		// Using built-in sort function Arrays.sort
		Arrays.sort(att, new Comparator<int[]>() {
 
			@Override
			// Compare values according to columns
			public int compare(final int[] entry1, final int[] entry2) {
 
				// To sort in descending order revert
				// the '>' Operator
				// if (entry1[col] > entry2[col])
				// return 1;
				// else //(entry1[col] >= entry2[col])
				// return -1;
				return new Integer(entry1[col]).compareTo(entry2[col]);
				// return entry1[col].
			}
		}); // End of function call sort().
	}
 
	static class pair {
		int V, I;
 
		pair(int v, int i) {
			V = v;
			I = i;
		}
	}
 
	public static int[] swap(int data[], int left, int right) {
		int temp = data[left];
		data[left] = data[right];
		data[right] = temp;
		return data;
	}
 
	public static int[] reverse(int data[], int left, int right) {
		while (left < right) {
			int temp = data[left];
			data[left++] = data[right];
			data[right--] = temp;
		}
		return data;
	}
 
	public static boolean findNextPermutation(int data[]) {
		if (data.length <= 1)
			return false;
 
		int last = data.length - 2;
		while (last >= 0) {
			if (data[last] < data[last + 1]) {
				break;
			}
			last--;
		}
		if (last < 0)
			return false;
 
		int nextGreater = data.length - 1;
		for (int i = data.length - 1; i > last; i--) {
			if (data[i] > data[last]) {
				nextGreater = i;
				break;
			}
		}
		data = swap(data, nextGreater, last);
		data = reverse(data, last + 1, data.length - 1);
		return true;
	}
 
	static long nCr(long n, long r) {
		if (n == r)
			return 1;
		return fact(n) / (fact(r) * fact(n - r));
	}
 
	static long fact(long n) {
		long res = 1;
		for (long i = 2; i <= n; i++)
			res = res * i;
		return res;
	}
 
	/*
	 * static boolean prime[] = new boolean[1000007];
	 * 
	 * public static void sieveOfEratosthenes(int n) {
	 * 
	 * for (int i = 0; i < n; i++) prime[i] = true; * for (int p = 2; p * p <=
	 * n; p++) { // If prime[p] is not changed, then it is a prime if (prime[p]
	 * == true) { // Update all multiples of p for (int i = p * p; i <= n; i +=
	 * p) prime[i] = false; } }
	 * 
	 * // Print all prime numbers // for(int i = 2; i <= n; i++) // { //
	 * if(prime[i] == true) // System.out.print(i + " "); // } }
	 */
 
	static long power(long fac2, int y, int p) {
		long res = 1;
		fac2 = fac2 % p;
		while (y > 0) {
			if (y % 2 == 1)
				res = (res * fac2) % p;
			fac2 = (fac2 * fac2) % p;
		}
		return res;
	}
 
	static long modInverse(long fac2, int p) {
		return power(fac2, p - 2, p);
	}
 
	static boolean isPrime(int n) {
		// Corner cases
		if (n <= 1)
			return false;
		if (n <= 3)
			return true;
 
		// This is checked so that we can skip
		// middle five numbers in below loop
		if (n % 2 == 0 || n % 3 == 0)
			return false;
 
		for (int i = 5; i * i <= n; i = i + 6)
			if (n % i == 0 || n % (i + 2) == 0)
				return false;
 
		return true;
	}
 
	public static BigInteger lcmm(String a, String b) {
		BigInteger s = new BigInteger(a);
		BigInteger s1 = new BigInteger(b);
		BigInteger mul = s.multiply(s1);
		BigInteger gcd = s.gcd(s1);
		BigInteger lcm = mul.divide(gcd);
		return lcm;
	}
 
}