Problem:
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Distance II

Your task is to calculate the distance between two $n$ dimensional vectors $x = \{x_1, x_2, ..., x_n\}$ and $y = \{y_1, y_2, ..., y_n\}$.

The Minkowski's distance defined below is a metric which is a generalization of both the Manhattan distance and the Euclidean distance.

\[
D_{xy} = (\sum_{i=1}^n |x_i - y_i|^p)^{\frac{1}{p}}
\]

It can be the Manhattan distance

\[
D_{xy} = |x_1 - y_1| + |x_2 - y_2| + ... + |x_n - y_n|
\]

where $p = 1 $.

It can be the Euclidean distance

\[
D_{xy} = \sqrt{(|x_1 - y_1|)^{2} + (|x_2 - y_2|)^{2} + ... + (|x_n - y_n|)^{2}}
\]

where $p = 2 $.

Also, it can be the Chebyshev distance

\[
D_{xy} = max_{i=1}^n (|x_i - y_i|)
\]

where $p = \infty$

Write a program which reads two $n$ dimensional vectors $x$ and $y$, and calculates Minkowski's distance where $p = 1, 2, 3, \infty$ respectively.

Input

In the first line, an integer $n$ is given. In the second and third line, $x = \{x_1, x_2, ... x_n\}$ and $y = \{y_1, y_2, ... y_n\}$ are given respectively. The elements in $x$ and $y$ are given in integers.

Output

Print the distance where $p = 1, 2, 3$ and $\infty$ in a line respectively.
The output should not contain an absolute error greater than 10-5.

Constraints

$1 \leq n \leq 100$

$0 \leq x_i, y_i \leq 1000$

Sample Input

3
1 2 3
2 0 4

Sample Output

4.000000
2.449490
2.154435
2.000000

Solution in JAVA (judge verdict: Wrong Answer)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
	public static void main(String[] args) {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
	    
		String x = "";
		try {
			x = in.readLine();
			int n = Integer.parseInt(x);

			String[] xs = in.readLine().split(" ");
			String[] ys = in.readLine().split(" ");

			int[] px = new int[xs.length];
			int[] py = new int[ys.length];
			for (int i = 0; i < px.length; i++) {
				px[i] = Integer.parseInt(xs[i]);
				py[i] = Integer.parseInt(ys[i]);
			}

			double ans = 0.0;
			for (int i = 0; i < px.length; i++) {
				ans = ans + Math.abs(px[i] - py[i]);
			}
			System.out.println(ans);

			ans = 0.0;
			for (int i = 0; i < px.length; i++) {
				ans = ans + (px[i] - py[i]) * (px[i] - py[i]);
			}
			System.out.println(Math.sqrt(ans));

			ans = 0.0;
			for (int i = 0; i < px.length; i++) {
				ans = ans + Math.abs(px[i] - py[i] * (px[i] - py[i]) * (px[i] - py[i]));
			}
			System.out.println(Math.cbrt(ans));

			ans = 0.0;
			for (int i = 0; i < px.length; i++) {
				double d = Math.abs(px[i] - py[i]);
				if (ans < d) ans = d;
			}
			System.out.println(ans);
			
		} catch (NumberFormatException e) {
			// TODO Auto-generated catch block
			e.printStackTrace();
		} catch (IOException e) {
			// TODO Auto-generated catch block
			e.printStackTrace();
		}
	}
}