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.

Constraints

* $1 \leq n \leq 100$
* $0 \leq x_i, y_i \leq 1000$

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.

Example

Input

3
1 2 3
2 0 4


Output

4.000000
2.449490
2.154435
2.000000
/*
String[]の初期化、整数の型がlongになっているか
関数（splitとかcontainsとか）内では"+"とかは正規表現となされるので"\\+"とする
配列のコピーはarr1=arr.clone()
HashSet<>[]はおのおの初期化した？？？？？
負の数の割り算は絶対値が小さいほうに切り捨てられるex.-1/2=0ので、ある値>kみたいな式だとバグらせやすいので注意
ある値>=kとして、切り上げた値をとるべき
toLowerCase()はアルファベット以外に対して行ってもエラーにはならない
 */
//入力終了→Ctrl+D
import java.util.*;
import java.awt.*;
import java.awt.Graphics.*;
import static java.lang.System.*;
import static java.lang.Math.*;
public class Main {
    public static void main(String[] $) {
        Scanner sc = new Scanner(in);
        while (sc.hasNext()){
            int n=sc.nextInt();
            double[] x=new double[n];
            double[] y=new double[n];
            for (int i = 0; i < n; i++) {
                x[i]=sc.nextDouble();
            }
            double l1=0,l2=0,l3=0,l8=0;
            for (int i = 0; i < n; i++) {
                y[i]=abs(sc.nextDouble()-x[i]);
                l1+=y[i];
                l2+=pow(y[i],2);
                l3+=pow(y[i],3);
                l8=max(l8,y[i]);
            }
            out.println(l1);
            out.println(sqrt(l2));
            out.println(pow(l3,(double) 1/3));
            out.println(l8);
        }
    }
}
