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:
// practice with rainboy
import java.io.*;
import java.util.*;

public class CF578C extends PrintWriter {
	CF578C() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF578C o = new CF578C(); o.main(); o.flush();
	}

	int[] aa;
	long cross(long x1, long y1, long x2, long y2) {
		return x1 * y2 - y1 * x2;
	}
	long cross3(int i, int j, int k) {
		return cross(j - i, aa[j] - aa[i], k - i, aa[k] - aa[i]);
	}
	double weakness(int i, int j, double x) {
		return Math.abs((aa[i] - x * i) - (aa[j] - x * j));
	}
	void main() {
		int n = sc.nextInt();
		aa = new int[n + 1];
		for (int i = 1; i <= n; i++) {
			aa[i] = sc.nextInt();
			aa[i] += aa[i - 1];
		}
		Integer[] ii = new Integer[n + 1];
		for (int h = 0; h <= n; h++)
			ii[h] = h;
		Arrays.sort(ii, 1, n + 1, (i, j) -> {
			long c = cross3(0, i, j);
			return c == 0 ? i - j : c < 0 ? -1 : 1;
		});
		int[] stack = new int[n + 2];
		int cnt = 0;
		stack[cnt++] = ii[0];
		stack[cnt++] = ii[1];
		for (int h = 2; h <= n; h++) {
			int i = ii[h];
			while (cnt >= 2 && cross3(stack[cnt - 2], stack[cnt - 1], i) >= 0)
				cnt--;
			stack[cnt++] = i;
		}
		stack[cnt] = 0;
		double ans = 3e9;
		for (int p = 0, q = 1; p < cnt; p++) {
			int ux = stack[p + 1] - stack[p];
			int uy = aa[stack[p + 1]] - aa[stack[p]];
			while (true) {
				int vx = stack[q + 1] - stack[q];
				int vy = aa[stack[q + 1]] - aa[stack[q]];
				if (cross(ux, uy, vx, vy) >= 0)
					break;
				q = (q + 1) % cnt;
			}
			ans = Math.min(ans, weakness(stack[p], stack[q], (double) uy / ux));
		}
		println(ans);
	}
}