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:
/*-
 * Codeforces problem :
 * http://codeforces.com/problemset/problem/645/G
 *
 * Find the number of distinct possible suffixes of length 2 or 3 for a
 * given string with minimum length 4 or more.
 * Two identical suffixes should not be appended. 
 * 
 */

import java.io.BufferedReader;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;

/**
 * @author Karthik Venkataraman
 * @email kafy83@gmail.com
 */
public class MaxDiffEuclideanDistance {

    static class InputReader {

        private static final int inputKb = 5 * 1024;
        private final BufferedReader bufferedReader;
        private StringTokenizer stringTokenizer;

        InputReader(InputStream stream) {
            bufferedReader = new BufferedReader(new InputStreamReader(stream), inputKb);
        }

        private String readNext() {
            while (stringTokenizer == null || !stringTokenizer.hasMoreTokens()) {
                try {
                    stringTokenizer = new StringTokenizer(bufferedReader.readLine());
                } catch (IOException ex) {
                    throw new IllegalArgumentException(ex);
                }
            }
            return stringTokenizer.nextToken();
        }

        private int nextInt() {
            return Integer.parseInt(readNext());
        }
    }

    static class Point {

        private double x;
        private double y;

        //w.r.t (a,0)
        private double xa;
        private double ya;
        private double da;
        private double ra;
        private double theta;

        Point(double x, double y, double a) {
            this.x = x;
            this.y = y;
            this.xa = (x - a) / 2;
            this.ya = y / 2;
            this.da = Math.sqrt(xa * xa + ya * ya);
            this.ra = Math.sqrt((xa + a) * (xa + a) + ya * ya);
            this.theta = Math.atan2(ya, xa);
        }
    }

    static class Angle implements Comparable<Angle> {

        private double angleInRad;
        private int pointIdx;

        public Angle(double angleInRad, int pointIdx) {
            this.angleInRad = angleInRad;
            this.pointIdx = pointIdx;
        }

        @Override
        public int hashCode() {
            int hash = 5;
            hash = 79 * hash + (int) (Double.doubleToLongBits(this.angleInRad) ^ (Double.doubleToLongBits(this.angleInRad) >>> 32));
            hash = 79 * hash + this.pointIdx;
            return hash;
        }

        @Override
        public boolean equals(Object obj) {
            if (obj == null) {
                return false;
            }
            if (getClass() != obj.getClass()) {
                return false;
            }
            final Angle other = (Angle) obj;
            if (Double.doubleToLongBits(this.angleInRad) != Double.doubleToLongBits(other.angleInRad)) {
                return false;
            }
            if (this.pointIdx != other.pointIdx) {
                return false;
            }
            return true;
        }

        public int getPointIdx() {
            return pointIdx;
        }

        @Override
        public int compareTo(Angle that) {
            Angle angle1 = this;
            Angle angle2 = that;
            double d1 = angle1.angleInRad;
            double d2 = angle2.angleInRad;
            if (d1 > d2) {
                return 1;
            }
            if (d2 > d1) {
                return -1;
            }
            long d1bits = Double.doubleToLongBits(d1);
            long d2bits = Double.doubleToLongBits(d2);
            return (d1bits == d2bits) ? 0 : (d1bits > d2bits) ? 1 : -1;
        }
    }

    static class AngleStack {

        private Angle[] stack;
        private int sz;

        AngleStack(int capacity) {
            stack = new Angle[capacity];
        }

        public void pushIfDifferentPoint(Angle x) {
            if (sz > 0 && stack[sz - 1].getPointIdx() == x.getPointIdx()) {
                stack[sz - 1] = null;
                sz--;
            } else {
                stack[sz++] = x;
            }
        }

        public void resetStack() {
            sz = 0;
        }

        public boolean isEmpty() {
            return sz == 0;
        }
    }

    private int a;
    private int n;
    private Angle[] angle;
    private AngleStack angleStack;
    private Point[] points;

    private double make2Quad(double d) {
        while (d >= Math.PI) {
            d -= 2 * Math.PI;
        }
        while (d < -Math.PI) {
            d += 2 * Math.PI;
        }
        return d;
    }

    private void input() {
        InputReader sc = new InputReader(System.in);
        n = sc.nextInt();
        points = new Point[n];
        angle = new Angle[2 * n];
        angleStack = new AngleStack(2 * n);
        a = sc.nextInt();
        for (int i = 0; i < n; i++) {
            points[i] = new Point(sc.nextInt(), sc.nextInt(), a);
        }
    }

    private void compute() {
        double lo = 0.0;
        double hi = 2 * a;
        while (hi - lo > Math.max(1, hi) * 1e-9) {
            double mid = (lo + hi) / 2;
            double rvar = mid / 2;
            int idx = 0;
            boolean isVeryLargeGuess = false;
            for (int i = 0; i < n; i++) {
                if (points[i].da + points[i].ra <= rvar) {
                    hi = mid;
                    isVeryLargeGuess = true;
                    break;
                }
                /*-
                 * distance between two polar coordinates (r1, theta1)
                 * and (r2, theta2)
                 * 
                 *  d = sqrt(r1^2+r2^2-2*r^1*r^2*cos(theta1-theta2))
                 *
                 *  ra^2 = da^2+rvar^2-2*da*rvar*cos(theta1-theta2))
                 *  
                 * cos (difference of polar angle between da and rvar) =
                 * (da^2+rvar^2-ra^2)/2*da*rvar
                 * 
                 * 
                 * polar angle of point + (difference of polar angle between da and rvar)
                 * polar angle of point - (difference of polar angle between da and rvar)
                 * 
                 */
                double cosaInRad = (points[i].da * points[i].da + rvar * rvar - points[i].ra * points[i].ra) / (2 * points[i].da * rvar);
                if (cosaInRad < 1 && cosaInRad > -1) {
                    double alphaDiffInRad = Math.acos(cosaInRad);
                    double thetaPlusAlpha = make2Quad(points[i].theta + alphaDiffInRad);
                    double thetaMinusAlpha = make2Quad(points[i].theta - alphaDiffInRad);
                    angle[idx++] = new Angle(thetaPlusAlpha, i);
                    angle[idx++] = new Angle(thetaMinusAlpha, i);
                }
            }
            if (isVeryLargeGuess) {
                continue;
            }
            angleStack.resetStack();
            Arrays.sort(angle, 0, idx);
            /*
             * Increase the guess if sorting angle puts  
             * same point's distance to a and -a next to each other.
             * else decrease the guess;
             */
            for (int i = 0; i < idx; i++) {
                angleStack.pushIfDifferentPoint(angle[i]);
            }
            if (angleStack.isEmpty()) {
                lo = mid;
            } else {
                hi = mid;
            }
        }
        System.out.println(String.format("%.15f", hi));
    }

    public static void main(String[] args) {
        MaxDiffEuclideanDistance maxDiffEuclideanDistance = new MaxDiffEuclideanDistance();
        maxDiffEuclideanDistance.input();
        maxDiffEuclideanDistance.compute();
    }
}