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.
import java.util.Scanner;

public class E {
    static int n;
    static double[] a;

    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);

        n = cin.nextInt();
        a = new double[n];
        for (int i=0;i<n;i++) {
            a[i]=cin.nextDouble();
        }

        double left, right;
        left = -1e5;
        right = 1e5;
        for (int i = 0; i < 100; i++) {
            double x = (left + right) / 2;
            if (f(x) > f((x + right) / 2)) {
                left = x;
            } else {
                right = (x + right) / 2;
            }
        }

        System.out.println(f(right));

        cin.close();
    }

    private static double f(double x) {
        double ans = 0;
        double sum = 0;
        for (int i = 0; i < n; i++) {
            sum += a[i] - x;
            if (sum > ans) ans = sum;
            if (sum < 0) sum = 0;
        }

        sum = 0;
        for (int i = 0; i < n; i++) {
            sum -= a[i] - x;
            if (sum > ans) ans = sum;
            if (sum < 0) sum = 0;
        }
        return ans;
    }
}
