Problem: Due to atheistic Soviet past, Christmas wasn't officially celebrated in Russia for most of the twentieth century. As a result, the Russian traditions for Christmas and New Year mixed into one event celebrated on the New Year but including the tree, a Santa-like 'Grandfather Frost', presents and huge family reunions and dinner parties all over the country. Bying a Tree at the New Year and installing it in the house is a tradition. Usually the whole family decorates the tree on the New Year Eve. We hope that Codeforces is a big and loving family, so in this problem we are going to decorate a tree as well.

So, our decoration consists of n pieces, each piece is a piece of colored paper, its border is a closed polyline of a special shape. The pieces go one by one as is shown on the picture. The i-th piece is a polyline that goes through points: (0, 0), (0, y0), (1, y1), (2, y2), ..., (k, yk), (k, 0). The width of each piece equals k.

<image> The figure to the left shows the decoration, the figure to the right shows the individual pieces it consists of.

The piece number 1 (shown red on the figure) is the outer piece (we see it completely), piece number 2 (shown yellow) follows it (we don't see it completely as it is partially closed by the first piece) and so on. The programmers are quite curious guys, so the moment we hung a decoration on the New Year tree we started to wonder: what area of each piece can people see?

Input

The first line contains two integers, n and k (1 ≤ n, k ≤ 300). Each of the following n lines contains k + 1 integers — the description of the polyline. If the i-th line contains ontegers yi, 0, yi, 1, ..., yi, k, that means that the polyline of the i-th piece goes through points (0, 0), (0, yi, 0), (1, yi, 1), (2, yi, 2), ..., (k, yi, k), (k, 0) (1 ≤ yi, j ≤ 1000).

Output

Print n real numbers — for each polyline, the area of its visible part.

The answer will be considered correct if its relative or absolute error do not exceed 10 - 4. 

Examples

Input

2 2
2 1 2
1 2 1


Output

3.000000000000
0.500000000000


Input

1 1
1 1


Output

1.000000000000


Input

4 1
2 7
7 2
5 5
6 4


Output

4.500000000000
1.250000000000
0.050000000000
0.016666666667

Solution (in JAVA):

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

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

class TaskE {
    public void solve(int testNumber, InputReader in, PrintWriter out) {
        int n = in.nextInt();
        int k = in.nextInt();
        int[][] y = new int[n][k + 1];
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j <= k; ++j) {
                y[i][j] = in.nextInt();
            }
        }
        double[] area = new double[n];
        for (int j = 0; j < k; ++j) {
            int[] left = new int[n];
            int[] right = new int[n];
            for (int i = 0; i < n; ++i) {
                left[i] = y[i][j];
                right[i] = y[i][j + 1];
            }
            updateArea(left, right, area);
        }
        for (double x : area) {
            out.println(String.format("%.20f", x));
        }
    }

    private void updateArea(int[] left, int[] right, double[] area) {
        int n = left.length;
        double[] x = new double[2 * n + 5];
        double[] y = new double[2 * n + 5];
        double[] nx = new double[2 * n + 5];
        double[] ny = new double[2 * n + 5];
        int cnt = 2;
        x[0] = 0;
        y[0] = 0;
        x[1] = 1;
        y[1] = 0;
        double ca = 0;
        for (int which = 0; which < n; ++which) {
            int l = left[which];
            int r = right[which];
            int ncnt = 0;
            int state = 0;
            for (int i = 0; i < cnt; ++i) {
                double cx = x[i];
                double cy = y[i];
                double wy = l + (r - l) * cx;
                if (wy < cy - 1e-10) {
                    if (state == 1) {
                        double min = x[i - 1];
                        double max = x[i];
                        while (max - min > 1e-15) {
                            double middle = (min + max) / 2;
                            double val = y[i - 1] + (y[i] - y[i - 1]) / (x[i] - x[i - 1]) * (middle - x[i - 1]);
                            double got = l + middle * (r - l);
                            if (got > val) {
                                min = middle;
                            } else {
                                max = middle;
                            }
                        }
                        nx[ncnt] = max;
                        ny[ncnt] = l + max * (r - l);
                        ++ncnt;
                        state = 2;
                    }
                    nx[ncnt] = cx;
                    ny[ncnt] = cy;
                    ++ncnt;
                } else if (state == 0) {
                    if (i == 0) {
                        nx[ncnt] = 0;
                        ny[ncnt] = l;
                        ++ncnt;
                    } else {
                        double min = x[i - 1];
                        double max = x[i];
                        while (max - min > 1e-15) {
                            double middle = (min + max) / 2;
                            double val = y[i - 1] + (y[i] - y[i - 1]) / (x[i] - x[i - 1]) * (middle - x[i - 1]);
                            double got = l + middle * (r - l);
                            if (got > val) {
                                max = middle;
                            } else {
                                min = middle;
                            }
                        }
                        nx[ncnt] = max;
                        ny[ncnt] = l + max * (r - l);
                        ++ncnt;
                    }
                    state = 1;
                }
            }
            if (state == 1) {
                nx[ncnt] = 1;
                ny[ncnt] = r;
                ++ncnt;
                state = 2;
            }
            double[] tmp = x;
            x = nx;
            nx = tmp;
            tmp = y;
            y = ny;
            ny = tmp;
            cnt = ncnt;
            double na = 0;
            for (int i = 0; i + 1 < cnt; ++i) {
                na += (x[i + 1] - x[i]) * (y[i] + y[i + 1]) / 2;
            }
            area[which] += na - ca;
            ca = na;
        }
    }
}

class InputReader {
    public BufferedReader reader;
    public StringTokenizer tokenizer;

    public InputReader(InputStream stream) {
        reader = new BufferedReader(new InputStreamReader(stream));
        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());
    }

    }