Problem: You are given a sequence of n integers a1, a2, ..., an. 

Determine a real number x such that the weakness of the sequence a1 - x, a2 - x, ..., an - x is as small as possible.

The weakness of a sequence is defined as the maximum value of the poorness over all segments (contiguous subsequences) of a sequence.

The poorness of a segment is defined as the absolute value of sum of the elements of segment.

Input

The first line contains one integer n (1 ≤ n ≤ 200 000), the length of a sequence.

The second line contains n integers a1, a2, ..., an (|ai| ≤ 10 000).

Output

Output a real number denoting the minimum possible weakness of a1 - x, a2 - x, ..., an - x. Your answer will be considered correct if its relative or absolute error doesn't exceed 10 - 6.

Examples

Input

3
1 2 3


Output

1.000000000000000


Input

4
1 2 3 4


Output

2.000000000000000


Input

10
1 10 2 9 3 8 4 7 5 6


Output

4.500000000000000

Note

For the first case, the optimal value of x is 2 so the sequence becomes  - 1, 0, 1 and the max poorness occurs at the segment "-1" or segment "1". The poorness value (answer) equals to 1 in this case. 

For the second sample the optimal value of x is 2.5 so the sequence becomes  - 1.5, - 0.5, 0.5, 1.5 and the max poorness occurs on segment "-1.5 -0.5" or "0.5 1.5". The poorness value (answer) equals to 2 in this case.

Solution (in JAVA):

Solution:
import java.util.*;
public class Main {
    static int n;
    static int[] arr;
	public static void main (String[] args){
	    Scanner sc = new Scanner(System.in);
	    n = sc.nextInt();
	    arr = new int[n];
	    for(int i = 0; i < n; ++i) arr[i] = sc.nextInt();
	    
	    //find x
	    double l = -(1e9 + 3), r = 1e9 + 3;
	    for(int i = 0; i < 200; ++i){
	        double g1 = (l * 2 + r) / 3;
	        double g2 = (l + r * 2) / 3;
	        
	        if(go(g1) > go(g2)) l = g1;
	        else r = g2;
	    }
	    
	    System.out.printf("%.8f%n", go(l));
	}
	static double go(double in){
	    double[] seq = new double[n];
	    for(int i = 0; i < n; ++i) seq[i] = arr[i] - in;
	    
	    double curr = 0, out = 0, min = 0, max = 0;
	    for(int i = 0; i < n; ++i){
	        curr += seq[i];
	        double temp1 = Math.abs(curr - min);
	        out = Math.max(out, temp1);
	        double temp2 = Math.abs(curr - max);
	        out = Math.max(out, temp2);
	        
	        min = Math.min(min, curr);
	        max = Math.max(curr, max);
	    }
	    
	    return out;
	}
}