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.*;

public class Main {
	static ArrayList<Integer> adj[];

	static int k;
	static int[] a;
	static int b[];
	static int m;

	static class Pair implements Comparable<Pair> {
		long a;
		long b;
		public Pair(long c, long l) {
			a = c;
			b = l;
		}
		@Override
		public int compareTo(Pair o) {
		if(o.a-o.b>this.a-this.b)
			return 1;
		else if(o.a-o.b==this.a-this.b)
			return 0;
		else {
			return -1;
		}
	}
}

	static ArrayList<Pair> adjlist[];
	// static char c[];
	static long mod = (long) (1e9+7);
	static int V;
	static long INF = (long) 1E16;
	static int n;
	static char c[];
	static Pair p[];
static int grid[][];
static Scanner sc = new Scanner(System.in);
static PrintWriter out = new PrintWriter(System.out);
public static void main(String args[]) throws Exception {
	int t=sc.nextInt();
	while(t-->0) {
	int n=sc.nextInt();
	double l=sc.nextDouble();
	TreeSet<Double> set=new TreeSet<>();
	double a[]=new double[n];
	for (int i = 0; i < a.length; i++) {
		a[i]=sc.nextDouble();
		set.add(a[i]);
	}
	double spda=1;
	double spdb=1;
	double starta=0.0;
	double startb=l;
	double meet=(l/2.0);
	double time=0.0;
	int z=0;
	while(true) {
		z++;
		double dist1=l;
		double dist2=l;
		if(set.higher(starta)!=null)
		 dist1=set.higher(starta)-starta;
		if(set.lower(startb)!=null)	
		 dist2=startb-set.lower(startb);
		//System.out.println((dist1/spda)+" "+(dist2/spdb));
		if((dist1/spda)<(dist2/spdb)&&(dist1/spda)<meet) {
			starta+=dist1;
			time+=(dist1/spda);
			
			startb-=(dist1/spda)*spdb;
			
			spda++;
		}
		else if((dist1/spda)>(dist2/spdb)&&(dist2/spdb)<meet) {
			startb-=dist2;
			time+=(dist2/spdb);	
			starta+=(dist2/spdb)*spda;
			
			spdb++;
			}
		else if((dist1/spda)==(dist2/spdb)&&(dist2/spdb)<meet) {
			startb-=dist2;
			spdb++;
			
			starta+=dist1;
			time+=(dist1/spda);
			
			spda++;
		}
		else if((dist2/spdb)<meet) {
			startb-=dist2;
			time+=(dist2/spdb);	
			starta+=(dist2/spdb)*spda;
			
			spdb++;
			
		}
		else if((dist1/spda)<meet) {
			starta+=dist1;
			time+=(dist1/spda);
			
			startb-=(dist1/spda)*spdb;
			
			spda++;
			
		}
	else {
		time+=meet;
		break;
		}
		meet=(startb-starta)/(spda+spdb);
	}
	System.out.println(time);
	
	}
	
	out.flush();
}


static void dfs(int u) {
	vis[u]=true;
	for(int v:adj[u]) {
		if(!vis[v]) {
			dfs(v);
		}
	}
}
static void update(int idx,int val) {
	if(idx==0) {
		total-=a[0];
		if(idx+1<a.length)
		total-=Math.max(0,a[idx+1]-a[idx]);
		a[0]=val;
		total+=a[0];
		if(idx+1<a.length)
		total+=Math.max(0,a[idx+1]-a[idx]);
		
		return;
	}
	if(idx>0) {
		total-=Math.max(0,a[idx]-a[idx-1]);
	}
	 if(idx+1<a.length) {
		total-=Math.max(0,a[idx+1]-a[idx]);
	}
	a[idx]=val;
	 if(idx>0) {
		total+=Math.max(0,a[idx]-a[idx-1]);
	}
	 if(idx+1<a.length) {
		total+=Math.max(0,a[idx+1]-a[idx]);
	}
}
static int[] compreessarray(int c[]) {
	Stack<Integer> st=new Stack<>();
	for (int i = 0; i < c.length; i++) {
		if(st.isEmpty()||c[i]!=st.peek())
		st.add(c[i]);
	}
	b=new int[st.size()];
	for (int i = b.length-1; i>=0; i--) {
		b[i]=st.pop();
	}
	return b;
}
static int div4[];
static long [] reversearray(long a[]) {
	for (int i = 0; i <a.length/2; i++) {
		long temp=a[i];
		a[i]=a[a.length-i-1];
		a[a.length-i-1]=temp;
	}
	return a;
}
static int [] reversearray(int a[]) {
	for (int i = 0; i <=a.length/2; i++) {
		int temp=a[i];
		a[i]=a[a.length-i-1];
		a[a.length-i-1]=temp;
	}
	return a;
}
static class SegmentTree {		// 1-based DS, OOP
	
	int N; 			//the number of elements in the array as a power of 2 (i.e. after padding)
	int[] array, sTree, lazy;
	
	SegmentTree(int[] in)		
	{
		array = in; N = in.length - 1;
		sTree = new int[N<<1];		//no. of nodes = 2*N - 1, we add one to cross out index zero
		lazy = new int[N<<1];
		build(1,1,N);
	}
	
	void build(int node, int b, int e)	// O(n)
	{
		if(b == e)					
			sTree[node] = array[b];
		else						
		{
			int mid = b + e >> 1;
			build(node<<1,b,mid);
			build(node<<1|1,mid+1,e);
			sTree[node] = sTree[node<<1]+sTree[node<<1|1];
		}
	}
	
	
	void update_point(int index, int val)			// O(log n)
	{
		index += N - 1;				
		sTree[index] += val;			
		while(index>1)				
		{
			index >>= 1;
			sTree[index] = sTree[index<<1] + sTree[index<<1|1];		
		}
	}
	
	
	void update_range(int i, int j, int val)		// O(log n) 
	{
		update_range(1,1,N,i,j,val);
	}
	
	void update_range(int node, int b, int e, int i, int j, int val)
	{
		if(i > e || j < b)		
			return;
		if(b >= i && e <= j)		
		{
			sTree[node] += (e-b+1)*val;			
			lazy[node] += val;				
		}							
		else		
		{
			int mid = b + e >> 1;
			propagate(node, b, mid, e);
			update_range(node<<1,b,mid,i,j,val);
			update_range(node<<1|1,mid+1,e,i,j,val);
			sTree[node] = sTree[node<<1] + sTree[node<<1|1];		
		}
		
	}
	void propagate(int node, int b, int mid, int e)		
	{
		lazy[node<<1] += lazy[node];
		lazy[node<<1|1] += lazy[node];
		sTree[node<<1] += (mid-b+1)*lazy[node];		
		sTree[node<<1|1] += (e-mid)*lazy[node];		
		lazy[node] = 0;
	}
	
	int query(int i, int j)
	{
		return query(1,1,N,i,j);
	}
	
	int query(int node, int b, int e, int i, int j)	// O(log n)
	{
		if(i>e || j <b)
			return 0;		
		if(b>= i && e <= j)
			return sTree[node];
		int mid = b + e >> 1;
		propagate(node, b, mid, e);
		int q1 = query(node<<1,b,mid,i,j);
		int q2 = query(node<<1|1,mid+1,e,i,j);
		return q1 + q2;	
				
	}
}
static  TreeSet<Integer> ts=new TreeSet();
static HashMap<Integer, Integer> compress(int a[]) {
		 ts = new TreeSet<>();
		HashMap<Integer, Integer> hm = new HashMap<>();
		for (int x : a)
			ts.add(x);
		for (int x : ts) {
			hm.put(x, hm.size() + 1);
		}
		return hm;
	}	

	static  class FenwickTree { // one-based DS

		int n;
		int[] ft;
		
		FenwickTree(int size) { n = size; ft = new int[n+1]; }
		
		int rsq(int b) //O(log n)
		{
			int sum = 0;
			while(b > 0) { sum += ft[b]; b -= b & -b;}		//min?
			return sum;
		}
		
		int rsq(int a, int b) { return rsq(b) - rsq(a-1); }	
		
		void point_update(int k, int val)	//O(log n), update = increment
		{
			while(k <= n) { ft[k] += val; k += k & -k; }		//min?
		}
	}
	static ArrayList<Integer> euler=new ArrayList<>();
	

	static ArrayList<Integer> arr;

	
static long total;
static TreeMap<Integer,Integer> map1;

		static int zz;
//static int dp(int idx,int left,int state) {
//if(idx>=k-((zz-left)*2)||idx+1==n) {
//	return 0;
//}
//if(memo[idx][left][state]!=-1) {
//	return memo[idx][left][state];
//}
//int ans=a[idx+1]+dp(idx+1,left,0);
//if(left>0&&state==0&&idx>0) {
//	ans=Math.max(ans,a[idx-1]+dp(idx-1,left-1,1));
//}
//return memo[idx][left][state]=ans;
//}21
static HashMap<Integer,Integer> map;
static int maxa=0;
static int ff=123123;

static int[][] memo;


	static long modmod=998244353;
	static int dx[]= {1,-1,0,0};
	 static int dy[]= {0,0,1,-1};	
	static class BBOOK implements Comparable<BBOOK>{
		int t;
		int alice;
		int bob;
		public BBOOK(int x,int y,int z) {
		t=x;
		alice=y;
		bob=z;
		}
		@Override
		public int compareTo(BBOOK o) {
			return this.t-o.t;
		}
	}
	private static long lcm(long a2, long b2) {
	return (a2*b2)/gcd(a2,b2);
	}
	static class Edge implements Comparable<Edge>
	{
		int node;long cost ; long time; 
		
		Edge(int a, long b,long c) { node = a;	cost = b; time=c; }
		
		public int compareTo(Edge e){ return Long.compare(time,e.time);	}
	}
	
        static void sieve(int N)	// O(N log log N) 
	{
		isComposite = new int[N+1];					
		isComposite[0] = isComposite[1] = 1;			// 0 indicates a prime number
		primes = new ArrayList<Integer>();
 
		for (int i = 2; i <= N; ++i) 					//can loop till i*i <= N if primes array is not needed O(N log log sqrt(N)) 
			if (isComposite[i] == 0) 					//can loop in 2 and odd integers for slightly better performance
			{
				primes.add(i);
				if(1l * i * i <= N)
					for (int j = i * i; j <= N; j += i)	// j = i * 2 will not affect performance too much, may alter in modified sieve
						isComposite[j] = 1;
			}   
	}
    static TreeSet<Integer> factors;
    static TreeSet<Integer> primeFactors(int N)		// O(sqrt(N) / ln sqrt(N))
	{
		 factors = new TreeSet<Integer>();		//take abs(N) in case of -ve integers
		int idx = 0, p = primes.get(idx);

		while(1l*p * p <= N)
		{
			while(N % p == 0) { factors.add(p); N /= p; }
			p = primes.get(++idx);
		}

		if(N != 1)						// last prime factor may be > sqrt(N)
			factors.add(N);				// for integers whose largest prime factor has a power of 1
		return factors;
	}

	static class UnionFind {
		int[] p, rank, setSize;
		int numSets;
		int max[];

		public UnionFind(int N) {
			p = new int[numSets = N];
			rank = new int[N];
			setSize = new int[N];
			for (int i = 0; i < N; i++) {
				p[i] = i;
				setSize[i] = 1;
			}
		}

		public int findSet(int i) {
			return p[i] == i ? i : (p[i] = findSet(p[i]));
		}

		public boolean isSameSet(int i, int j) {
			return findSet(i) == findSet(j);
		}
		public int chunion(int i,int j, int x2) {
			if (isSameSet(i, j))
				return 0;
			numSets--;
			int x = findSet(i), y = findSet(j);
			int z=findSet(x2);
			p[x]=z;;
			p[y]=z;
		return x;
		}

		public void unionSet(int i, int j) {
			if (isSameSet(i, j))
				return;
			numSets--;
			int x = findSet(i), y = findSet(j);
			if (rank[x] > rank[y]) {
				p[y] = x;
				setSize[x] += setSize[y];


			} else {
				p[x] = y;
				setSize[y] += setSize[x];
				if (rank[x] == rank[y])
					rank[y]++;


			}
		}

		

		public int numDisjointSets() {
			return numSets;
		}

		public int sizeOfSet(int i) {
			return setSize[findSet(i)];
		}
	}


	static class Quad implements Comparable<Quad> {
		int u;
		int v;
		char state;
		int turns;

		public Quad(int i, int j, char c, int k) {
			u = i;
			v = j;
			state = c;
			turns = k;
		}

		public int compareTo(Quad e) {
			return (int) (turns - e.turns);
		}

	}

	static long manhatandistance(long x, long x2, long y, long y2) {
		return Math.abs(x - x2) + Math.abs(y - y2);
	}

	static long fib[];

	static long fib(int n) {
		if (n == 1 || n == 0) {
			return 1;
		}
		if (fib[n] != -1) {
			return fib[n];
		} else
			return fib[n] = ((fib(n - 2) % mod + fib(n - 1) % mod) % mod);
	}

	static class Point  implements Comparable<Point>{
		long x, y;

		Point(long counth, long counts) {
			x = counth;
			y = counts;
		}

		@Override
		public int compareTo(Point p )
		{
			return Long.compare(p.y*1l*x, p.x*1l*y);
		}
	}

	static TreeSet<Long> primeFactors(long N) // O(sqrt(N) / ln sqrt(N))
	{
		TreeSet<Long> factors = new TreeSet<Long>(); // take abs(N) in case of -ve integers
		int idx = 0, p = primes.get(idx);

		while (p * p <= N) {
			while (N % p == 0) {
				factors.add((long) p);
				N /= p;
			}
			if (primes.size() > idx + 1)
				p = primes.get(++idx);
			else
				break;
		}

		if (N != 1) // last prime factor may be > sqrt(N)
			factors.add(N); // for integers whose largest prime factor has a power of 1
		return factors;
	}

	static boolean visited[];

	/**
	 * static int bfs(int s) { Queue<Integer> q = new LinkedList<Integer>();
	 * q.add(s); int count=0; int maxcost=0; int dist[]=new int[n]; dist[s]=0;
	 * while(!q.isEmpty()) {
	 * 
	 * int u = q.remove(); if(dist[u]==k) { break; } for(Pair v: adj[u]) {
	 * maxcost=Math.max(maxcost, v.cost);
	 * 
	 * 
	 * 
	 * if(!visited[v.v]) {
	 * 
	 * visited[v.v]=true; q.add(v.v); dist[v.v]=dist[u]+1; maxcost=Math.max(maxcost,
	 * v.cost); } }
	 * 
	 * } return maxcost; }
	 **/

	static boolean[] vis2;

	static boolean f2 = false;

	static long[][] matMul(long[][] a2, long[][] b, int p, int q, int r) // C(p x r) = A(p x q) x (q x r) -- O(p x q x
																			// r)
	{
		long[][] C = new long[p][r];
		for (int i = 0; i < p; ++i) {
			for (int j = 0; j < r; ++j) {
				for (int k = 0; k < q; ++k) {
					C[i][j] = (C[i][j] + (a2[i][k] % mod * b[k][j] % mod)) % mod;
					C[i][j] %= mod;
				}

			}
		}
		return C;
	}
	public static int[] schuffle(int[] a2) {
		for (int i = 0; i < a2.length; i++) {
			int x =  (int) (Math.random() * a2.length);
			int temp = a2[x];
			a2[x] = a2[i];
			a2[i] = temp;
		}
		return a2;
	}


	static boolean vis[];
	static HashSet<Integer> set = new HashSet<Integer>();

	static long modPow(long ways, long count, long mod) // O(log e)
	{
		ways %= mod;
		long res = 1;
		while (count > 0) {
			if ((count & 1) == 1)
				res = (res * ways) % mod;
			ways = (ways * ways) % mod;
			count >>= 1;
		}
		return res % mod;
	}

	static long gcd(long l, long o) {
		if (o == 0) {
			return l;
		}
		return gcd(o, l % o);
	}

	static int[] isComposite;
	static int[] valid;

	static ArrayList<Integer> primes;
	static ArrayList<Integer> l1;

	
	static TreeSet<Integer> primus = new TreeSet<Integer>();
	
	static void sieveLinear(int N)
	{
		int[] lp = new int[N + 1];								//lp[i] = least prime divisor of i
		for(int i = 2; i <= N; ++i)
		{
			if(lp[i] == 0)
			{
				primus.add(i);
				lp[i] = i;
			}
			int curLP = lp[i];
			for(int p: primus)
				if(p > curLP || p * i > N)
					break;
				else
					lp[p * i] = i;
		}
	}
	

	public static long[] schuffle(long[] a2) {
		for (int i = 0; i < a2.length; i++) {
			int x =  (int) (Math.random() * a2.length);
			long temp = a2[x];
			a2[x] = a2[i];
			a2[i] = temp;
		}
		return a2;
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}

		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public char nextChar() throws IOException {
			return next().charAt(0);
		}

		public Long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}

		public int[] nxtArr(int n) throws IOException {
			int[] ans = new int[n];
			for (int i = 0; i < n; i++)
				ans[i] = nextInt();
			return ans;
		}
		
	}
	public static int[] sortarray(int a[]) {
		schuffle(a);
		Arrays.sort(a);
		return a;
		}
		public static long[] sortarray(long a[]) {
			schuffle(a);
			Arrays.sort(a);
			return a;
		}
		public static int[] readarray(int n) throws IOException {
			int a[]=new int[n];
			for (int i = 0; i < a.length; i++) {
			a[i]=sc.nextInt();
			}
			return a;
		}
		public static long[] readlarray(int n) throws IOException {
			long a[]=new long[n];
			for (int i = 0; i < a.length; i++) {
			a[i]=sc.nextLong();
			}
			return a;
		}
}