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 PYTHON3):

Solution:
from bisect import *
from collections import *
from math import *
from heapq import *
from typing import List
from itertools import *
from operator import *
from functools import *

#------------------------------------------------------------------------
import os
import sys
from io import BytesIO, IOBase
# region fastio
 
BUFSIZE = 8192
 
class FastIO(IOBase):
    newlines = 0
 
    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
 
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
 
    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)
 
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")
 
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#------------------------------------------------------------------------
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
#------------------------------------------------------------------------

@lru_cache(None)
def fact(x):
    if x<2:
        return 1
    return fact(x-1)*x

@lru_cache(None)
def per(i,j):
    return fact(i)//fact(i-j)

@lru_cache(None)
def com(i,j):
    return per(i,j)//fact(j)

def linc(f,t,l,r):
    while l<r:
        mid=(l+r)//2
        if t>f(mid):
            l=mid+1
        else:
            r=mid
    return l

def rinc(f,t,l,r):
    while l<r:
        mid=(l+r+1)//2
        if t<f(mid):
            r=mid-1
        else:
            l=mid
    return l

def ldec(f,t,l,r):
    while l<r:
        mid=(l+r)//2
        if t<f(mid):
            l=mid+1
        else:
            r=mid
    return l

def rdec(f,t,l,r):
    while l<r:
        mid=(l+r+1)//2
        if t>f(mid):
            r=mid-1
        else:
            l=mid
    return l

def isprime(n):
    for i in range(2,int(n**0.5)+1):
        if n%i==0:
            return False
    return True

def binfun(x):
    c=1
    res=0
    for w in arr:
        res+=w
        if res>x:
            res=w
        c+=1
    return c

def lowbit(n):
    return n&-n

class BIT:
    def __init__(self,arr):
        self.arr=arr
        self.n=len(arr)-1
        
    def update(self,x,v):
        while x<=self.n:
            self.arr[x]+=v
            x+=x&-x

    def query(self,x):
        ans=0
        while x:
            ans+=self.arr[x]
            x&=x-1
        return ans

class smt:
    def __init__(self,l,r,arr):
        self.l=l
        self.r=r
        self.value=(1<<31)-1 if l<r else arr[l]
        mid=(l+r)//2
        if(l<r):
            self.left=smt(l,mid,arr)
            self.right=smt(mid+1,r,arr)
            self.value&=self.left.value&self.right.value
        #print(l,r,self.value)
    def setvalue(self,x,val):
        if(self.l==self.r):
            self.value=val
            return
        mid=(self.l+self.r)//2
        if(x<=mid):
            self.left.setvalue(x,val)
        else:
            self.right.setvalue(x,val)
        self.value=self.left.value&self.right.value
    def ask(self,l,r):
        if(l<=self.l and r>=self.r):
            return self.value
        val=(1<<31)-1
        mid=(self.l+self.r)//2
        if(l<=mid):
            val&=self.left.ask(l,r)
        if(r>mid):
            val&=self.right.ask(l,r)
        return val

class UFS:
    def __init__(self,n):
        self.parent=[i for i in range(n)]
        self.ranks=[0]*n

    def find(self,x):
        if x!=self.parent[x]:
            self.parent[x]=self.find(self.parent[x])
        return self.parent[x]

    def union(self,u,v):
        pu,pv=self.find(u),self.find(v)
        if pu==pv:
            return False
        if self.ranks[pu]>=self.ranks[pv]:
            self.parent[pv]=pu
            if self.ranks[pv]==self.ranks[pu]:
                self.ranks[pu]+=1
        else:
            self.parent[pu]=pv

def Prime(n):
    c=0
    prime=[]
    flag=[0]*(n+1)
    for i in range(2,n+1):
        if not flag[i]:
            prime.append(i)
            c+=1
        for j in range(c):
            if i*prime[j]>n: break
            flag[i*prime[j]]=prime[j]
            if i%prime[j]==0: break
    return prime

def dij(s,graph):
    d={}
    d[s]=0
    heap=[(0,s)]
    seen=set()
    while heap:
        dis,u=heappop(heap)
        if u in seen:
            continue
        for v in graph[u]:
            if v not in d or d[v]>d[u]+graph[u][v]:
                d[v]=d[u]+graph[u][v]
                heappush(heap,(d[v],v))
    return d

'''
import time
s=time.time()
for i in range(2000):
    print(0)
e=time.time()
print(e-s)
'''

t=N()
for i in range(t):
    n,l=RL()
    a=RLL()
    a1=[0]*(n+2)
    a2=[0]*(n+2)
    for i in range(1,n+2):
        if i==1:
            tmp=a[0]    
        elif i==n+1:
            tmp=(l-a[-1])/(n+1)
        else:
            tmp=(a[i-1]-a[i-2])/i
        a1[i]=a1[i-1]+tmp
    for i in range(n,-1,-1):
        if i==n:
            tmp=l-a[n-1]
        elif i==0:
            tmp=a[0]/(n+1)
        else:
            tmp=(a[i]-a[i-1])/(n-i+1)
        a2[i]=a2[i+1]+tmp
    #print(a1,a2)
    for i in range(1,n+2):
        if a1[i]>=a2[i]:
            break
    if i==1:
        b,c=0,a[0]
    elif i==n+1:
        b,c=a[-1],l
    else:
        b,c=a[i-2],a[i-1]
    x=((c-b)/(n-i+2)+a2[i]-a1[i-1])/(1/i+1/(n-i+2))
    ans=a1[i-1]+x/i
    print(ans)


'''
n,m=RL()
arr1=[]
arr2=[]
for i in range(n):
    arr1.append(RLL())
for i in range(m):
    arr2.append(RLL())
x=[0]*(10**6+2)
for i in range(n):
    for j in range(m):
        a,b=arr1[i]
        c,d=arr2[j]
        if a>c or b>d:
            continue
        x[c-a+1]=max(x[c-a+1],d-b+1)
y=0
ans=inf
for i in range(10**6+1,-1,-1):
    ans=min(ans,i+y)
    y=max(y,x[i])
print(ans)
'''