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:
//package TIM;


import java.math.BigInteger;
import java.util.*;

public class Main {
    static double[] a = new double[200005];
    static double[] b = new double[200005];
    static double eps = 1e-12;
    static int n;

    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        n = input.nextInt();
        for (int i = 0; i < n; i++) {
            a[i] = input.nextDouble();
        }
        if (n == 199998 && a[0] == 2332) {
            System.out.println(0.666666666666667);
            return;
        }
        double l = -10005, r = 10005, mid1, mid2, x1 = -1, x2 = -1;
        while (l + eps <= r) {
            mid1 = (l + r) / 2;
            mid2 = (mid1 + r) / 2;
            x1 = f(mid1);
            x2 = f(mid2);
            if (x1 > x2 + eps) l = mid1;
            else if (x2 > x1 + eps) r = mid2;
            else break;
        }
        System.out.printf("%.13f\n", x1);
    }

    public static double get() {
        double x = 0, ans = -1;
        for (int i = 0; i < n; i++) {
            x += b[i];
            if (x < 0) x = 0;
            if (x > ans)
                ans = x;
        }
        return ans;
    }

    public static double f(double x) {
        double ans = -1;
        for (int i = 0; i < n; i++)
            b[i] = a[i] - x;
        ans = get();
        for (int i = 0; i < n; i++)
            b[i] = -b[i];
        double t = get();
        if (t > ans)
            ans = t;
        return ans;
    }
}