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 WeaknessPoorness3 {

    public static final double ERR = 1e-7;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
        }

        double result = ternarySearch(a);
        System.out.println(result);
    }

    private static double ternarySearch(int[] a) {
        double min = -10000;
        double max = 10000;
        double w1 = 0;
        double w2 = 0;

        while (Math.abs(max - min) > ERR || Math.abs(w1-w2)>ERR ) {
            double m1 = min + (max - min) / 3;
            double m2 = min + 2 * (max - min) / 3;
            w1 = weakness(a, m1);
            w2 = weakness(a, m2);
            if (w1 < w2) {
                max = m2;
            } else {
                min = m1;
            }
        }

        return weakness(a, (max + min) / 2);
    }

    private static double weakness(int[] a, double x) {
        double maxSum = Integer.MIN_VALUE;
        double minSum = Integer.MAX_VALUE;

        double curMax = 0;
        double curMin = 0;
        for (int anA : a) {
            curMax += (anA - x);
            curMin += (anA - x);
            if (curMax > maxSum) {
                maxSum = curMax;
            }
            if (curMin < minSum) {
                minSum = curMin;
            }
            if (curMax < 0) {
                curMax = 0;
            }
            if (curMin > 0) {
                curMin = 0;
            }
        }
        return Math.max(Math.abs(minSum), Math.abs(maxSum));
    }
}