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.*;
import java.io.*;
import static java.lang.System.in;

public class Main {
    static double[] a,b;
    static int n;
    public static void main(String[] args)throws IOException {
        Scanner sc = new Scanner(System.in);
        n = sc.nextInt();
        a = new double[n]; b = new double[n];
        for(int i=0;i<n;i++) a[i]=sc.nextDouble();
        double hi = 10000.0, lo = -10000.0;
        for(int pass=0;pass<60;pass++){
            double mid = (hi+lo)/2;
            double[] cur = check(mid);
            if(cur[0]>cur[1]) lo = mid; // positive part is too much. We let b[i] = a[i]-mid, so needs more mid
            else hi = mid;
        }
        double[] res = check((lo+hi)/2);
        double ans = Math.max(Math.abs(res[0]),Math.abs(res[1]));
        System.out.println(ans);
    }
    static double help(double[] c){
        double ret = 0,cur = 0;
        for(int i=0;i<n;i++){
            cur += c[i];
            ret = Math.max(ret,cur);
            if(cur<0) cur = 0;
        }
        return ret;
    }
    static double[] check(double mid){
        for(int i=0;i<n;i++) b[i] = a[i]-mid;
        double[] ans = new double[2];
        ans[0] = help(b);
        for(int i=0;i<n;i++) b[i]*=-1;
        ans[1] = Math.abs(help(b));
        return ans;
    }
}