Problem: After a drawn-out mooclear arms race, Farmer John and the Mischievous Mess Makers have finally agreed to establish peace. They plan to divide the territory of Bovinia with a line passing through at least two of the n outposts scattered throughout the land. These outposts, remnants of the conflict, are located at the points (x1, y1), (x2, y2), ..., (xn, yn).

In order to find the optimal dividing line, Farmer John and Elsie have plotted a map of Bovinia on the coordinate plane. Farmer John's farm and the Mischievous Mess Makers' base are located at the points P = (a, 0) and Q = ( - a, 0), respectively. Because they seek a lasting peace, Farmer John and Elsie would like to minimize the maximum difference between the distances from any point on the line to P and Q.

Formally, define the difference of a line <image> relative to two points P and Q as the smallest real number d so that for all points X on line <image>, |PX - QX| ≤ d. (It is guaranteed that d exists and is unique.) They wish to find the line <image> passing through two distinct outposts (xi, yi) and (xj, yj) such that the difference of <image> relative to P and Q is minimized.

Input

The first line of the input contains two integers n and a (2 ≤ n ≤ 100 000, 1 ≤ a ≤ 10 000) — the number of outposts and the coordinates of the farm and the base, respectively.

The following n lines describe the locations of the outposts as pairs of integers (xi, yi) (|xi|, |yi| ≤ 10 000). These points are distinct from each other as well as from P and Q.

Output

Print a single real number—the difference of the optimal dividing line. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6. 

Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.

Examples

Input

2 5
1 0
2 1


Output

7.2111025509


Input

3 6
0 1
2 5
0 -3


Output

0.0000000000

Note

In the first sample case, the only possible line <image> is y = x - 1. It can be shown that the point X which maximizes |PX - QX| is (13, 12), with <image>, which is <image>.

In the second sample case, if we pick the points (0, 1) and (0, - 3), we get <image> as x = 0. Because PX = QX on this line, the minimum possible difference is 0.

Solution (in JAVA):

Solution:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskG solver = new TaskG();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskG {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int n = in.nextInt();
            double a = in.nextInt();
            Point[] p = new Point[n];
            for (int i = 0; i < n; ++i) {
                p[i] = new Point();
                p[i].x = in.nextInt();
                p[i].y = in.nextInt();
            }
            for (int i = 0; i < n; ++i) {
                double x1 = (p[i].x - a) / 2;
                double y1 = p[i].y / 2;
                double r1 = Math.sqrt((x1 + a) * (x1 + a) + y1 * y1);
                double d = Math.sqrt(x1 * x1 + y1 * y1);
                p[i].x1 = x1;
                p[i].y1 = y1;
                p[i].r1 = r1;
                p[i].d = d;
                p[i].polar = Math.atan2(y1, x1);
            }
            double left = 0.0;
            double right = 2 * a;
            Event[] events = new Event[2 * n];
            for (int i = 0; i < events.length; ++i) {
                events[i] = new Event();
            }
            int[] stack = new int[2 * n];
            outer:
            while (right - left > Math.max(1, right) * 1e-9) {
                double mid = (left + right) / 2;
                double r0 = mid / 2;
                int nev = 0;
                for (int i = 0; i < n; ++i) {
                    Point pp = p[i];
                    double r1 = pp.r1;
                    double d = pp.d;
                    if (d + r1 <= r0) {
                        right = mid;
                        continue outer;
                    }
                    double cosa = (r0 * r0 + d * d - r1 * r1) / (2 * r0 * d);
                    if (cosa < 1 && cosa > -1) {
                        double alpha = Math.acos(cosa);
                        double a1 = norm(pp.polar + alpha);
                        double a2 = norm(pp.polar - alpha);
                        events[nev].alpha = a1;
                        events[nev].what = i;
                        ++nev;
                        events[nev].alpha = a2;
                        events[nev].what = i;
                        ++nev;
                    }
                }
                Arrays.sort(events, 0, nev);
                int sp = 0;
                for (int i = 0; i < nev; ++i) {
                    int cur = events[i].what;
                    if (sp > 0 && stack[sp - 1] == cur) --sp;
                    else stack[sp++] = cur;
                }
                if (sp == 0) {
                    left = mid;
                } else {
                    right = mid;
                }
            }
            out.println(String.format("%.15f", right));
        }

        private double norm(double v) {
            while (v >= Math.PI) v -= 2 * Math.PI;
            while (v < -Math.PI) v += 2 * Math.PI;
            return v;
        }

        static class Point {
            double x;
            double y;
            double x1;
            double y1;
            double r1;
            double polar;
            double d;

        }

        static class Event implements Comparable<Event> {
            double alpha;
            int what;


            public int compareTo(Event o) {
                return Double.compare(alpha, o.alpha);
            }

        }

    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

    }
}