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.*;

public class Main {
    public static void main(String[] args) {
        Solver solver = new Solver();
        solver.solve();
    }
}

class Solver {
    Scanner stdin = new Scanner(System.in);
    
    final int MAXN = 200010;
    int[] a = new int[MAXN];
    final double INF = 1e6;
    final double EPS = 1e-8;
    final int TIME = 100;
    
    double poorness(int n, double x) {
        double sum = 0.0;
        double max = 0.0; // careful
        double min = 0.0; // careful
        double res = 0.0;
        for (int i = 1; i <= n; i++) {
            sum += a[i] - x;
            res = Math.max(res, Math.abs(sum - max));
            res = Math.max(res, Math.abs(sum - min));
            max = Math.max(max, sum);
            min = Math.min(min, sum);
        }
        return res;
    }
    
    void solve() {
        int n = stdin.nextInt();
        for (int i = 1; i <= n; i++)
            a[i] = stdin.nextInt();
        double L = -INF;
        double R = INF;
        for (int i = 0; i < TIME; i++) {
            double len = (R - L) / 3.0;
            double p = L + len;
            double q = R - len;
            if (poorness(n, p) > poorness(n, q))
                L = p;
            else R = q;
        }
        System.out.println(poorness(n, L));
    }
}