Problem: Brothers Fred and George Weasley once got into the sporting goods store and opened a box of Quidditch balls. After long and painful experiments they found out that the Golden Snitch is not enchanted at all. It is simply a programmed device. It always moves along the same trajectory, which is a polyline with vertices at the points (x0, y0, z0), (x1, y1, z1), ..., (xn, yn, zn). At the beginning of the game the snitch is positioned at the point (x0, y0, z0), and then moves along the polyline at the constant speed vs. The twins have not yet found out how the snitch behaves then. Nevertheless, they hope that the retrieved information will help Harry Potter and his team in the upcoming match against Slytherin. Harry Potter learned that at the beginning the game he will be at the point (Px, Py, Pz) and his super fast Nimbus 2011 broom allows him to move at the constant speed vp in any direction or remain idle. vp is not less than the speed of the snitch vs. Harry Potter, of course, wants to catch the snitch as soon as possible. Or, if catching the snitch while it is moving along the polyline is impossible, he wants to hurry the Weasley brothers with their experiments. Harry Potter catches the snitch at the time when they are at the same point. Help Harry.

Input

The first line contains a single integer n (1 ≤ n ≤ 10000). The following n + 1 lines contain the coordinates xi, yi, zi, separated by single spaces. The coordinates of any two consecutive points do not coincide. The next line contains the velocities vp and vs, the last line contains Px, Py, Pz, separated by single spaces. All the numbers in the input are integers, their absolute value does not exceed 104. The speeds are strictly positive. It is guaranteed that vs ≤ vp.

Output

If Harry Potter can catch the snitch while it is moving along the polyline (including the end (xn, yn, zn)), print "YES" in the first line (without the quotes). Print in the second line t, which is the earliest moment of time, when Harry will be able to catch the snitch. On the third line print three numbers X, Y, Z, the coordinates of the point at which this happens. The absolute or relative error in the answer should not exceed 10 - 6. If Harry is not able to catch the snitch during its moving along the described polyline, print "NO".

Examples

Input

4
0 0 0
0 10 0
10 10 0
10 0 0
0 0 0
1 1
5 5 25


Output

YES
25.5000000000
10.0000000000 4.5000000000 0.0000000000


Input

4
0 0 0
0 10 0
10 10 0
10 0 0
0 0 0
1 1
5 5 50


Output

NO


Input

1
1 2 3
4 5 6
20 10
1 2 3


Output

YES
0.0000000000
1.0000000000 2.0000000000 3.0000000000

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

/**
 * Codeforces #65C "Harry Potter and the Golden Snitch"
 * @link http://codeforces.ru/problemset/problem/65/C
 * @author Caesar
 */
public class Solution65C implements Runnable {

    private static final int EXIT_CODE = -1;

    private StringTokenizer tok;
    private BufferedReader in;
    private PrintWriter out;

    public static void main(String[] args) {
        new Thread(new Solution65C()).start();
    }

    public void run() {
        in = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        long start = System.currentTimeMillis();
        try {
            solve();
        } catch (Throwable ex) {
            ex.printStackTrace();
            System.exit(EXIT_CODE);
        } finally {
            out.flush();
        }
        System.err.println("Time used: " + (System.currentTimeMillis() - start));
    }

    private String readString() throws IOException {
        while (tok == null || !tok.hasMoreTokens()) {
            tok = new StringTokenizer(in.readLine());
        }
        return tok.nextToken();
    }

    private int readInt() throws IOException {
        return Integer.parseInt(readString());
    }

    private double distSqr(double x1, double y1, double z1, double x2, double y2, double z2) {
        return (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2) + (z1 - z2) * (z1 - z2);
    }

    private double dist(double x1, double y1, double z1, double x2, double y2, double z2) {
        return Math.sqrt(distSqr(x1, y1, z1, x2, y2, z2));
    }

    public void solve() throws IOException{
        final double eps = 1.0E-9;
        int n = readInt() + 1;
        double[] x = new double[n];
        double[] y = new double[n];
        double[] z = new double[n];
        for (int i = 0; i < n; ++i) {
            x[i] = readInt();
            y[i] = readInt();
            z[i] = readInt();
        }
        double vp = readInt();
        double vs = readInt();
        double xp = readInt();
        double yp = readInt();
        double zp = readInt();
        double t0 = 0.0;
        double vp2 = vp * vp;
        double vs2 = vs * vs;
        double vpvs2 = vp2 - vs2;
        for (int i = 1; i < n; ++i) {
            double xa = x[i - 1];
            double ya = y[i - 1];
            double za = z[i - 1];
            double xb = x[i];
            double yb = y[i];
            double zb = z[i];
            double rab2 = distSqr(xa, ya, za, xb, yb, zb);
            double rap2 = distSqr(xa, ya, za, xp, yp, zp);
            double rab = Math.sqrt(rab2);
            double rap = Math.sqrt(rap2);
            if (rap <= t0 * vp) {
                out.println("YES");
                out.println(t0);
                out.print(xa);
                out.print(' ');
                out.print(ya);
                out.print(' ');
                out.println(za);
                return;
            }
            double a = rab2 * vpvs2;
            double b = 2.0 * vs * (t0 * rab * vp2 + vs *
                    ((xp - xa) * (xb - xa) + (yp - ya) * (yb - ya) + (zp - za) * (zb - za)));
            double c = vs2 * (t0 * t0 * vp2 - rap2);
            if (Math.abs(a) < eps) {
                double s = -c / b;
                if (s >= 0 && s <= 1) {
                    double xq = xa + s * (xb - xa);
                    double yq = ya + s * (yb - ya);
                    double zq = za + s * (zb - za);
                    out.println("YES");
                    out.println(dist(xp, yp, zp, xq, yq, zq) / vp);
                    out.print(xq);
                    out.print(' ');
                    out.print(yq);
                    out.print(' ');
                    out.println(zq);
                    return;
                }
            } else {
                double d = b * b - 4.0 * a * c;
                if (d >= 0) {
                    double s1 = (-b - Math.sqrt(d)) / (2.0 * a);
                    double s2 = (-b + Math.sqrt(d)) / (2.0 * a);
                    if (s1 > s2) {
                        double temp = s1;
                        s1 = s2;
                        s2 = temp;
                    }
                    if (s1 >= 0 && s1 <= 1) {
                        double xq = xa + s1 * (xb - xa);
                        double yq = ya + s1 * (yb - ya);
                        double zq = za + s1 * (zb - za);
                        out.println("YES");
                        out.println(dist(xp, yp, zp, xq, yq, zq) / vp);
                        out.print(xq);
                        out.print(' ');
                        out.print(yq);
                        out.print(' ');
                        out.println(zq);
                        return;
                    }
                    if (s2 >= 0 && s2 <= 1) {
                        double xq = xa + s2 * (xb - xa);
                        double yq = ya + s2 * (yb - ya);
                        double zq = za + s2 * (zb - za);
                        out.println("YES");
                        out.println(dist(xp, yp, zp, xq, yq, zq) / vp);
                        out.print(xq);
                        out.print(' ');
                        out.print(yq);
                        out.print(' ');
                        out.println(zq);
                        return;
                    }
                }
            }
            t0 += rab / vs;
        }
        out.println("NO");
    }

}