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.*;

class Main{
  public static void main(String[] args) throws IOException{
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    int n;
    n = Integer.parseInt(br.readLine());
    String[] str1 = br.readLine().split(" ");
    String[] str2 = br.readLine().split(" ");
    int[] x = new int[n];
    int[] y = new int[n];
    double sum_2 = 0;
    double sum_3 = 0;
    double dist_1 = 0;
    double dist_2 = 0;
    double dist_3 = 0;
    double dist_m = 0;
    for (int i = 0; i < n; i++){
      x[i] = Integer.parseInt(str1[i]);
      y[i] = Integer.parseInt(str2[i]);
    }

    // p = 1
    for (int i = 0; i < n; i++){
      dist_1 += Math.abs(x[i]-y[i]);
    }
    // p = 2
    for (int i = 0; i < n; i++){
      sum_2 += Math.abs((x[i]-y[i])*(x[i]-y[i]));
    }
    dist_2 = Math.sqrt(sum_2);
    //p = 3
    for (int i = 0; i < n; i++){
      sum_3 += Math.abs(Math.pow((x[i]-y[i]),3));
    }
    dist_3 = Math.cbrt(sum_3);
    //p = ∞
    int max = x[0]-y[0];
    for (int i = 0; i < n; i++){
      if (x[i]-y[i] >= max)
        max = x[i]-y[i];
    }
    dist_m = max;
    System.out.println(dist_1);
    System.out.println(dist_2);
    System.out.println(dist_3);
    System.out.println(dist_m);

  }
}