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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.StringTokenizer;


public class WeaknessAndPoorness {
    
    static int v[];
    
    public static void main(String[] args)throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        v = new int[n];
        StringTokenizer st = new StringTokenizer(br.readLine() );
        for(int i=0; i<n; i++){
            v[i] = Integer.parseInt(st.nextToken() );
        }
        
        
        double min, max, m1, m2;
        min = -10010;
        max = 10010;
        
        while( max - min > 1e-11 ){
            m1 = min+(max-min)/3.;
            m2 = min+2*(max-min)/3.;
            
            if( calc(m1) < calc(m2)){
                max = m2;
            }else{
                min = m1;
            }
        }
        System.out.println(calc( (max+min)/2 ));
        
    }
    
    public static double calc( double x ){
        double ans = 0;
        double max, min;
        max = v[0]-x;
        min = v[0]-x;
        ans = Math.max(Math.abs(max), Math.abs(min));
        for(int i=1; i<v.length; i++){
            max = Math.max( max+v[i]-x, v[i]-x);
            ans = Math.max(ans, Math.abs(max));
            
            min = Math.min( min+v[i]-x, v[i]-x);
            ans = Math.max(ans, Math.abs(min));
        }
        
        return ans;
    }
}