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.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
        Scanner scanner = new Scanner(System.in);
        int N = scanner.nextInt();
        long[] a = new long[N];
        for(int i = 0; i < N; i++) a[i] = scanner.nextLong();

        double l = -1e5;
        double r = 1e5;
        for(int k = 0; k < 100; k++) {
            double mid1 = l + (r - l) / 3;
            double mid2 = l + ((r - l) / 3) * 2;

            if (calc(a, mid1) < calc(a, mid2)) r = mid2; else l = mid1;
        }

        System.out.println(calc(a, (l + r) / 2));

    
    }
    
    public static double calc(long[] a, double x) {
        double bestMax = -1e12;
        double bestMin = 1e12;
        double current = 0;
        for(int i = 0; i < a.length; i++) {
            current = Math.max(current + a[i] - x, 0);
            bestMax = Math.max(bestMax, current);
        }
        current = 0;
        for(int i = 0; i < a.length; i++) {
            current = Math.min(current + a[i] - x, 0);
            bestMin = Math.min(bestMin, current);
        }

        return Math.max(Math.abs(bestMin), bestMax);
    }
}