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 C (judge verdict: Time Limit Exceeded)

#include<stdio.h>
#include<math.h>
int main(){

  int i,n,score[1001],sum;
  double ave,sig,v;

  
  while(1){

    scanf("%d", &n );
    if ( n == 0 ) break;
    
    for ( i  = 0 ; i < n ; i++ ){
      scanf("%d", &score[i] );
    }

    sum = 0;
    
    for ( i = 0 ; i < n ; i++ ){
      sum = sum + score[i];
    }
    
    ave = 1.0 * sum / n;

    sig = 0;
    
    for ( i = 0 ; i < n ; i++ ){
      sig = sig + ( score[i] - ave ) * ( score[i] - ave );
    }

    v = sqrt( sig / n );

    printf("%.8lf\n", v );

  }
  
  return 0;
    
}