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.*;
import java.io.*;
public class WeaknessPoorness {
	static int n;
	static double[] arr, sum;
	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		n = Integer.parseInt(br.readLine());
		arr = new double[n];
		StringTokenizer st = new StringTokenizer(br.readLine());
		sum = new double[n];
		for(int i = 0; i < n; i++) {
			arr[i] = Integer.parseInt(st.nextToken());
			sum[i] = (i == 0 ? 0 : sum[i-1]) + arr[i];
		}
		double a = -30000; double b = 30000; int iters = 0;
		for(; iters < 100; iters++){
			double c = (a+a+b)/3; double d = (a+b+b)/3;
			if(f(c) > f(d)) a = c;
			else b = d;
		}
		System.out.println(f(a));
	}
	static double f(double x){
		double res = Math.max(calc(x,-1), calc(x,1)); 
		return res;
	}
	static double calc(double x, int sign){
		double max = Long.MIN_VALUE;
		int c = -1;
		for(int i = 0; i < n; i++){
			if((sum[i]-(c<0?0:sum[c])-x*(i-c))*sign < 0) c = i;
			double curr = (sum[i]-(c<0?0:sum[c])-x*(i-c))*sign;
			max = Double.max(curr, max);
		}
		return max;
	}
}