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.*;

public class Solution {

    BufferedReader br;
    PrintWriter out;
    StringTokenizer st;
    boolean eof;

    static final int Mod = 1000000007;
    static final double inf = 10000000000.0;

    double calcMax(double f[]) {
        double res = 0, cur = 0;
        for (int i = 0; i < f.length; i++) {
            cur = Math.max(f[i], cur + f[i]);
            res = Math.max(res, cur);
        }
        return res;
    }

    double calcMin(double f[]) {
        double res = 0, cur = 0;
        for (int i = 0; i < f.length; i++) {
            cur = Math.min(f[i], cur + f[i]);
            res = Math.min(res, cur);
        }
        return res;
    }

    void solve() throws IOException {
        int n = nextInt();
        double a[] = new double[n];
        double b[] = new double[n];
        for (int i = 0; i < n; i++) {
            a[i] = nextDouble();
        }
        double l = -inf, r = inf;
        for (int it = 0; it < 200; it++) {
            double x = (l + r) / 2;
            for (int i = 0; i < n; i++) {
                b[i] = a[i] - x;
            }
            double max = calcMax(b), min = calcMin(b);
            if (-min > max) r = x;
            else l = x;
        }
        for (int i = 0; i < n; i++) {
            b[i] = a[i] - l;
        }
        double res = calcMax(b);
        out.println(res);
    }

    Solution() throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        out.close();
    }

    public static void main(String[] args) throws IOException {
        new Solution();
    }

    String nextToken() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                eof = true;
                return null;
            }
        }
        return st.nextToken();
    }

    String nextString() {
        try {
            return br.readLine();
        } catch (IOException e) {
            eof = true;
            return null;
        }
    }

    int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }
}