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.io.*;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.math.MathContext;;

public class E {
	static BufferedReader br;
	static PrintWriter pw;
	static int inf = (int) 1e9;
	static long mod = (long) 1e9 + 7;

	static ArrayList<Integer> primes; // generated by sieve

	/*
	 * 1. Generating a list of prime factors of N
	 */
	static ArrayList<Long> primeFactors(Long N) // O(sqrt(N) / ln sqrt(N))
	{
		ArrayList<Long> factors = new ArrayList<Long>(); // take abs(N) in case of -ve integers
		int idx = 0, p = primes.get(idx);

		while (p * p <= N) {
			if (N % p == 0)
				factors.add((long) p);
			while (N % p == 0) {
				N /= p;
			}
			if (idx + 1 == primes.size())
				break;
			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 int[] isComposite;

	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 String reString(String s) {
		StringBuilder be = new StringBuilder();
		char[] se = s.toCharArray();
		for (int i = se.length - 1; i >= 0; i--)
			be.append(se[i]);
		return be.toString();
	}

	static int n;

	public static void main(String[] args) throws NumberFormatException, IOException, InterruptedException {
		br = new BufferedReader(new InputStreamReader(System.in));
		pw = new PrintWriter(System.out);

		StringTokenizer st = new StringTokenizer(br.readLine());
		int t = Integer.parseInt(st.nextToken());
		while (t-- > 0) {
			st = new StringTokenizer(br.readLine());
			int n = Integer.parseInt(st.nextToken());
			l = Integer.parseInt(st.nextToken());
			int[] a = nxtarr();
			double s = 0;
			double e = 1e9;
			double ans = -1;
//			System.out.println(Arrays.toString(a));
//			check(a, 1);
			for (int i = 0; i < 100; i++) {
				double mid = (s + e) / 2;
				if (check(a, mid)) {
					ans = mid;
					s = mid;
				} else
					e = mid;
			}
			pw.println(ans);
		}

		pw.flush();
		pw.close();

	}

	static int l;

	static boolean check(int[] a, double t) {
		double s = 0;
		int v = 1;
		double te = t;
		for (int i = 0; i < a.length; i++) {
			double d = a[i] - s;
			double time = d / v;
			if (time <= te) {
				te -= time;
				v++;
				s = a[i];
			} else {
				break;
			}
		}
		s += v * te;
		////
		double s2 = l;
		int v2 = 1;
		te = t;
		for (int i = a.length - 1; i >= 0; i--) {
			double d = s2 - a[i];
			double time = d / v2;
			if (time <= te) {
				te -= time;
				v2++;
				s2 = a[i];
			} else {
				break;
			}
		}
		s2 -= v2 * te;
		return s <= s2;
	}

	static int m, k;

	static double[] memo;
	static ArrayList<Integer>[] adj;
	static ArrayList<Integer> g;

	static long gcd(long a, long b) {
		if (a == 0)
			return b;
		return gcd(b % a, a);
	}

	// method to return LCM of two numbers
	static long lcm(long a, long b) {
		return (a * b) / gcd(a, b);
	}

	static double dis(int x, int y, int z, int w) {
		return Math.sqrt((x - z) * (x - z) + (y - w) * (y - w));
	}

	static int[] nxtarr() throws IOException {
		StringTokenizer st = new StringTokenizer(br.readLine());
		int[] a = new int[st.countTokens()];
		for (int i = 0; i < a.length; i++) {
			a[i] = Integer.parseInt(st.nextToken());
		}
		return a;
	}

	static class pair implements Comparable<pair> {
		int x;
		int y;

		public pair(int d, int u) {
			x = d;
			y = u;
		}

		@Override
		public int compareTo(pair o) {
			// TODO Auto-generated method stub
			if (y == o.y)
				return x - o.x;
			return y - o.y;
		}

	}

	static class triple implements Comparable<triple> {
		int x;
		int y;
		int z;

		public triple(int a, int b, int c) {
			x = a;
			y = b;
			z = c;
		}

		@Override
		public int compareTo(triple o) {
			// TODO Auto-generated method stub
			return x - o.x;
		}
	}

	static class LineSegment {

		Point p, q;

		LineSegment(Point a, Point b) {
			p = a;
			q = b;
		}

		boolean intersect(LineSegment ls) {
			Line l1 = new Line(p, q), l2 = new Line(ls.p, ls.q);
			if (l1.parallel(l2)) {
				if (l1.same(l2))
					return p.between(ls.p, ls.q) || q.between(ls.p, ls.q) || ls.p.between(p, q) || ls.q.between(p, q);
				return false;
			}
			Point c = l1.intersect(l2);
			return c.between(p, q) && c.between(ls.p, ls.q);
		}

	}

	static class Line {

		static final double INF = 1e9, EPS = 1e-9;

		double a, b, c;

		Line(Point p, Point q) {
			if (Math.abs(p.x - q.x) < EPS) {
				a = 1;
				b = 0;
				c = -p.x;
			} else {
				a = (p.y - q.y) / (q.x - p.x);
				b = 1.0;
				c = -(a * p.x + p.y);
			}

		}

		Line(Point p, double m) {
			a = -m;
			b = 1;
			c = -(a * p.x + p.y);
		}

		boolean parallel(Line l) {
			return Math.abs(a - l.a) < EPS && Math.abs(b - l.b) < EPS;
		}

		boolean same(Line l) {
			return parallel(l) && Math.abs(c - l.c) < EPS;
		}

		Point intersect(Line l) {
			if (parallel(l))
				return null;
			double x = (b * l.c - c * l.b) / (a * l.b - b * l.a);
			double y;
			if (Math.abs(b) < EPS)
				y = -l.a * x - l.c;
			else
				y = -a * x - c;

			return new Point(x, y);
		}

		Point closestPoint(Point p) {
			if (Math.abs(b) < EPS)
				return new Point(-c, p.y);
			if (Math.abs(a) < EPS)
				return new Point(p.x, -c);
			return intersect(new Line(p, 1 / a));
		}

	}

	static class Point implements Comparable<Point> {

		static final double EPS = 1e-9;

		double x, y;

		Point(double a, double b) {
			x = a;
			y = b;
		}

		public int compareTo(Point p) {
			if (Math.abs(x - p.x) > EPS)
				return x > p.x ? 1 : -1;
			if (Math.abs(y - p.y) > EPS)
				return y > p.y ? 1 : -1;
			return 0;
		}

		public double dist(Point p) {
			return Math.sqrt(sq(x - p.x) + sq(y - p.y));
		}

		double sq(double x) {
			return x * x;
		}

		Point rotate(double angle) {
			double c = Math.cos(angle), s = Math.sin(angle);
			return new Point(x * c - y * s, x * s + y * c);
		}
		// for integer points and rotation by 90 (counterclockwise) : swap x and y,
		// negate x

		Point rotate(double theta, Point p) // rotate around p
		{
			Vector v = new Vector(p, new Point(0, 0));
			return translate(v).rotate(theta).translate(v.reverse());
		}

		Point translate(Vector v) {
			return new Point(x + v.x, y + v.y);
		}

		Point reflectionPoint(Line l) // reflection point of p on line l
		{
			Point p = l.closestPoint(this);
			Vector v = new Vector(this, p);
			return this.translate(v).translate(v);
		}

		boolean between(Point p, Point q) {
			return x < Math.max(p.x, q.x) + EPS && x + EPS > Math.min(p.x, q.x) && y < Math.max(p.y, q.y) + EPS
					&& y + EPS > Math.min(p.y, q.y);
		}

		// returns true if it is on the line defined by a and b
		boolean onLine(Point a, Point b) {
			if (a.compareTo(b) == 0)
				return compareTo(a) == 0;
			return Math.abs(new Vector(a, b).cross(new Vector(a, this))) < EPS;
		}

		boolean onSegment(Point a, Point b) {
			if (a.compareTo(b) == 0)
				return compareTo(a) == 0;
			return onRay(a, b) && onRay(b, a);
		}

		// returns true if it is on the ray whose start point is a and passes through b
		boolean onRay(Point a, Point b) {
			if (a.compareTo(b) == 0)
				return compareTo(a) == 0;
			return new Vector(a, b).normalize().equals(new Vector(a, this).normalize()); // implement equals()
		}

		// returns true if it is on the left side of Line pq
		// add EPS to LHS if on-line points are accepted
		public static boolean ccw(Point p, Point q, Point r) {
			return new Vector(p, q).cross(new Vector(p, r)) > 0;
		}

		static boolean collinear(Point p, Point q, Point r) {
			return Math.abs(new Vector(p, q).cross(new Vector(p, r))) < EPS;
		}

		static double angle(Point a, Point o, Point b) // angle AOB
		{
			Vector oa = new Vector(o, a), ob = new Vector(o, b);
			return Math.acos(oa.dot(ob) / Math.sqrt(oa.norm2() * ob.norm2()));
		}

		double distToLine(Point p, Point a, Point b) // distance between point p and a line defined by points a, b (a !=
														// b)
		{
			if (a.compareTo(b) == 0)
				return p.dist(a);
			// formula: c = a + u * ab
			Vector ap = new Vector(a, p), ab = new Vector(a, b);
			double u = ap.dot(ab) / ab.norm2();
			Point c = a.translate(ab.scale(u));
			return p.dist(c);
		}
		// Another way: find closest point and calculate the distance between it and p

		double distToLineSegment(Point p, Point a, Point b) {
			Vector ap = new Vector(a, p), ab = new Vector(a, b);
			double u = ap.dot(ab) / ab.norm2();
			if (u < 0.0)
				return p.dist(a);
			if (u > 1.0)
				return p.dist(b);
			return distToLine(p, a, b);
		}
		// Another way : find closest point and calculate the distance between it and p
	}

	static class Polygon {
		// Cases to handle: collinear points, polygons with n < 3

		static final double EPS = 1e-9;

		Point[] g; // first point = last point, counter-clockwise representation

		Polygon(Point[] o) {
			g = o;
		}

		double perimeter() {
			double sum = 0.0;
			for (int i = 0; i < g.length - 1; ++i)
				sum += g[i].dist(g[i + 1]);
			return sum;
		}

		double area() // clockwise/anti-clockwise check, for convex/concave polygons
		{
			double area = 0.0;
			for (int i = 0; i < g.length - 1; ++i)
				area += g[i].x * g[i + 1].y - g[i].y * g[i + 1].x;
			return Math.abs(area) / 2.0; // negative value in case of clockwise
		}

		boolean isConvex() {
			if (g.length <= 3) // point or line
				return false;
			boolean ccw = Point.ccw(g[g.length - 2], g[0], g[1]); // edit ccw check to accept collinear points
			for (int i = 1; i < g.length - 1; ++i)
				if (Point.ccw(g[i - 1], g[i], g[i + 1]) != ccw)
					return false;
			return true;
		}

		boolean inside(Point p) // for convex/concave polygons - winding number algorithm
		{
			double sum = 0.0;
			for (int i = 0; i < g.length - 1; ++i) {
				double angle = Point.angle(g[i], p, g[i + 1]);
				if ((Math.abs(angle) < EPS || Math.abs(angle - Math.PI) < EPS) && p.between(g[i], g[i + 1]))
					return true;
				if (Point.ccw(p, g[i], g[i + 1]))
					sum += angle;
				else
					sum -= angle;
			}

			return Math.abs(2 * Math.PI - Math.abs(sum)) < EPS; // abs makes it work for clockwise
		}
		/*
		 * Another way if the polygon is convex 1. Triangulate the poylgon through p 2.
		 * Check if sum areas == poygon area 3. Handle empty polygon
		 */

		Polygon cutPolygon(Point a, Point b) // returns the left part of the polygon, swap a & b for the right part
		{
			Point[] ans = new Point[g.length << 1];
			Line l = new Line(a, b);
			Vector v = new Vector(a, b);

			int size = 0;
			for (int i = 0; i < g.length; ++i) {
				double left1 = v.cross(new Vector(a, g[i]));
				double left2 = i == g.length - 1 ? 0 : v.cross(new Vector(a, g[i + 1]));

				if (left1 + EPS > 0)
					ans[size++] = g[i];
				if (left1 * left2 + EPS < 0)
					ans[size++] = l.intersect(new Line(g[i], g[i + 1]));
			}

			if (size != 0 && ans[0] != ans[size - 1]) // necessary in case g[0] is not in the new polygon
				ans[size++] = ans[0];
			return new Polygon(Arrays.copyOf(ans, size));
		}

		static Polygon convexHull(Point[] points) // all points are unique, remove duplicates, edit ccw to accept
													// collinear points
		{
			int n = points.length;
			Arrays.sort(points);
			Point[] ans = new Point[n << 1];
			int size = 0, start = 0;

			for (int i = 0; i < n; i++) {
				Point p = points[i];
				while (size - start >= 2 && !Point.ccw(ans[size - 2], ans[size - 1], p))
					--size;
				ans[size++] = p;
			}
			start = --size;

			for (int i = n - 1; i >= 0; i--) {
				Point p = points[i];
				while (size - start >= 2 && !Point.ccw(ans[size - 2], ans[size - 1], p))
					--size;
				ans[size++] = p;
			}
			// if(size < 0) size = 0 for empty set of points
			return new Polygon(Arrays.copyOf(ans, size));
		}

		Point centroid() // center of mass
		{
			double cx = 0.0, cy = 0.0;
			for (int i = 0; i < g.length - 1; i++) {
				double x1 = g[i].x, y1 = g[i].y;
				double x2 = g[i + 1].x, y2 = g[i + 1].y;

				double f = x1 * y2 - x2 * y1;
				cx += (x1 + x2) * f;
				cy += (y1 + y2) * f;
			}
			double area = area(); // remove abs
			cx /= 6.0 * area;
			cy /= 6.0 * area;
			return new Point(cx, cy);
		}
	}

	static class Vector {

		double x, y;

		Vector(double a, double b) {
			x = a;
			y = b;
		}

		Vector(Point a, Point b) {
			this(b.x - a.x, b.y - a.y);
		}

		Vector scale(double s) {
			return new Vector(x * s, y * s);
		} // s is a non-negative value

		double dot(Vector v) {
			return (x * v.x + y * v.y);
		}

		double cross(Vector v) {
			return x * v.y - y * v.x;
		}

		double norm2() {
			return x * x + y * y;
		}

		Vector reverse() {
			return new Vector(-x, -y);
		}

		Vector normalize() {
			double d = Math.sqrt(norm2());
			return scale(1 / d);
		}
	}
}