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.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class E {

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());
		String[] split = br.readLine().split(" ");
		double[] numbers = new double[n];
		for (int i = 0; i < numbers.length; i++) {
			numbers[i] = Double.parseDouble(split[i]);
		}
		double binaryMin = -10000;
		double binaryMax = 10000;
		double result = 0;
		double lastResult = 28091994;
		while (Math.abs(lastResult - result) > 1e-7) {
			// System.out.println("binarymin" + binaryMin);
			// System.out.println("binarymax" + binaryMax);
			double pivot = (binaryMax + binaryMin) / 2;
			// System.out.println("PIVOT " + pivot);
			lastResult = result;
			result = f(n, numbers, pivot);
			// System.out.println("result" + result);
			// System.out.println("-----------");
			if (result > 0) {
				binaryMax = pivot;
			} else {
				binaryMin = pivot;
			}
		}
		System.out.println(Math.abs(result));

	}

	private static double f(int n, double[] numbers, double x) {
		double max = 0.0;
		double min = 0.0;
		double imin = 0.0;
		double imax = 0.0;

		for (int i = 0; i < numbers.length; i++) {
			double number = numbers[i] + x;
			double newMax = imax + number;
			if (newMax > 0) {
				imax = newMax;
			} else {
				imax = 0;
			}

			double newMin = imin + number;
			if (newMin < 0) {
				imin = newMin;
			} else {
				imin = 0;
			}
			if (newMax > max) {
				max = newMax;
			}
			if (newMin < min) {
				min = newMin;
			}
		}
		double restaDeAbs = Math.abs(max) - Math.abs(min);
		// System.out.println("restadeabs" + restaDeAbs);
		// System.out.println("absmax" + Math.abs(max));
		// System.out.println("absmin" + Math.abs(min));
		// if (Math.abs(restaDeAbs) < 1e-9) {
		// return 0;
		// }
		return (restaDeAbs > 0) ? Math.abs(max) : -Math.abs(min);
	}

}
