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.InputStreamReader;
import java.math.*;
import java.text.*;
public class Main {
    public static void main(String[] args) throws Exception {
        // Here your code !
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        NumberFormat nf = NumberFormat.getInstance();
        nf.setMaximumFractionDigits(6);
        nf.setGroupingUsed(false);
        int vector = Integer.parseInt(br.readLine());
        String[] tmpx1 = br.readLine().split(" ");
        String[] tmpy1 = br.readLine().split(" ");
        double[] vecx1 = new double[vector];
        double[] vecy1 = new double[vector];
        for(int i=0;i<tmpx1.length;i++)
            vecx1[i] = Double.parseDouble(tmpx1[i]);
        for(int i=0;i<tmpy1.length;i++)
            vecy1[i] = Double.parseDouble(tmpy1[i]);
        double p1 = 0;//p1の答え
        double p2 = 0;
        double p3 = 0;
        double pin = 0;
        //p1
        for(int i=0;i<vector;i++){
            p1 += Math.abs(vecx1[i]-vecy1[i]);
        }
        //p2
        for(int i=0;i<vector;i++){
            p2 += Math.pow(Math.abs(vecx1[i]-vecy1[i]),2);
        }
        p2 = Math.sqrt(p2);
        //p3
        for(int i=0;i<vector;i++){
            p3 += Math.pow(Math.abs(vecx1[i]-vecy1[i]),3);
        }
        p3 = Math.cbrt(p3);
        pin = Math.sqrt(p1);
        System.out.println(nf.format(p1));
        System.out.println(nf.format(p2));
        System.out.println(nf.format(p3));
        System.out.println(nf.format(pin));
    }
}