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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.IOException;
import java.io.UncheckedIOException;
import java.math.BigDecimal;
import java.io.Closeable;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.io.InputStream;

/**
 * @author rozerhacks
 */
public class Main {
    public static void main(String[] args) throws Exception {
        Thread thread = new Thread(null, new TaskAdapter(), "rozerhacks", 1 << 29);
        thread.start();
        thread.join();
    }

    static class TaskAdapter implements Runnable {
        @Override
        public void run() {
            InputStream inputStream = System.in;
            OutputStream outputStream = System.out;
            FastInput in = new FastInput(inputStream);
            FastOutput out = new FastOutput(outputStream);
            CDiscreteAcceleration solver = new CDiscreteAcceleration();
            int testCount = Integer.parseInt(in.next());
            for (int i = 1; i <= testCount; i++)
                solver.solve(i, in, out);
            out.close();
        }
    }

    static class CDiscreteAcceleration {
        public void solve(int testNumber, FastInput in, FastOutput out) {
            int n = in.readInt();
            int l = in.readInt();
            int[] ar = new int[n];
            for (int i = 0; i < n; i++) ar[i] = in.readInt();
            double timeL = 0;
            double timeR = 1000000000;
            int count = 0;
            double ans = -1;
            while (count++ < 200) {
                double mid = (timeL + timeR) / 2;
                double leftD = 0;
                double leftMid = mid;
                for (int i = 0; i < n; i++) {
                    double dist = ar[i] - leftD;
                    double time = dist / (i + 1);
                    if (time <= leftMid) {
                        leftMid -= time;
                        leftD = ar[i];
                    } else {
                        leftD += leftMid * (i + 1);
                        leftMid = 0;
                        break;
                    }
                }
                if (leftMid > 0) leftD += (leftMid * (n + 1));
                double rightD = l;
                double rightMid = mid;
                for (int i = n - 1; i >= 0; i--) {
                    double dist = rightD - ar[i];
                    double time = dist / (n - i);
                    if (time <= rightMid) {
                        rightMid -= time;
                        rightD = ar[i];
                    } else {
                        rightD -= rightMid * (n - i);
                        rightMid = 0;
                        break;
                    }
                }
                if (rightMid > 0) rightD -= (rightMid * (n + 1));
                if (leftD >= rightD) {
                    ans = mid;
                    timeR = mid - (1e-9);
                } else {
                    timeL = mid + (1e-9);
                }
            }
            out.println(ans);
        }

    }

    static class FastInput {
        private final InputStream is;
        private final StringBuilder defaultStringBuf = new StringBuilder(1 << 13);
        private final byte[] buf = new byte[1 << 13];
        private int bufLen;
        private int bufOffset;
        private int next;

        public FastInput(InputStream is) {
            this.is = is;
        }

        private int read() {
            while (bufLen == bufOffset) {
                bufOffset = 0;
                try {
                    bufLen = is.read(buf);
                } catch (IOException e) {
                    bufLen = -1;
                }
                if (bufLen == -1) {
                    return -1;
                }
            }
            return buf[bufOffset++];
        }

        public void skipBlank() {
            while (next >= 0 && next <= 32) {
                next = read();
            }
        }

        public String next() {
            return readString();
        }

        public int readInt() {
            int sign = 1;

            skipBlank();
            if (next == '+' || next == '-') {
                sign = next == '+' ? 1 : -1;
                next = read();
            }

            int val = 0;
            if (sign == 1) {
                while (next >= '0' && next <= '9') {
                    val = val * 10 + next - '0';
                    next = read();
                }
            } else {
                while (next >= '0' && next <= '9') {
                    val = val * 10 - next + '0';
                    next = read();
                }
            }

            return val;
        }

        public String readString(StringBuilder builder) {
            skipBlank();

            while (next > 32) {
                builder.append((char) next);
                next = read();
            }

            return builder.toString();
        }

        public String readString() {
            defaultStringBuf.setLength(0);
            return readString(defaultStringBuf);
        }

    }

    static class FastOutput implements AutoCloseable, Closeable, Appendable {
        private final Writer os;
        private final StringBuilder cache = new StringBuilder(5 << 20);

        public FastOutput(Writer os) {
            this.os = os;
        }

        public FastOutput(OutputStream os) {
            this(new OutputStreamWriter(os));
        }

        public FastOutput append(CharSequence csq) {
            cache.append(csq);
            return this;
        }

        public FastOutput append(CharSequence csq, int start, int end) {
            cache.append(csq, start, end);
            return this;
        }

        public FastOutput append(char c) {
            cache.append(c);
            return this;
        }

        public FastOutput append(double c) {
            cache.append(new BigDecimal(c).toPlainString());
            return this;
        }

        public FastOutput println(double c) {
            return append(c).println();
        }

        public FastOutput println() {
            cache.append(System.lineSeparator());
            return this;
        }

        public FastOutput flush() {
            try {
                os.append(cache);
                os.flush();
                cache.setLength(0);
            } catch (IOException e) {
                throw new UncheckedIOException(e);
            }
            return this;
        }

        public void close() {
            flush();
            try {
                os.close();
            } catch (IOException e) {
                throw new UncheckedIOException(e);
            }
        }

        public String toString() {
            return cache.toString();
        }

    }
}