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.Scanner;

public class weakpoor {
	public static void main(String[] args) {
		Scanner scan=new Scanner(System.in);
		int n=scan.nextInt();
		double[] a=new double[n];
		for(int i=0;i<n;i++) a[i]=scan.nextInt();
		
		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(a,g1) > go(a,g2)) l = g1;
	        else r = g2;
	    }
	    
	    System.out.println(go(a,l));
	    
	    
	}
	public static double go(double[] a, double x) {
		double max=0, min=0;
		double maxStreak=0, minStreak=0;
		
		for(int i=0;i<a.length;i++) {
			maxStreak+=a[i]-x;
			minStreak+=a[i]-x;
			
			if(maxStreak<0) maxStreak=0;
			if(minStreak>0) minStreak=0;
			
			max=Math.max(max,maxStreak);
			min=Math.min(min,minStreak);
		}
		return Math.max(max,-min);
	}
}