579_E. Weakness and Poorness
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.
```python
def main():
	input()
	a = list(map(int, input().split()))
	
	def f(a):
	    maxend = maxnow = 0
	    for x in a:
	        maxend = max(0, maxend + x)
	        maxnow = max(maxnow, maxend)
	    return maxnow
	
	f1 = lambda x: f(i - x for i in a)
	f2 = lambda x: f(x - i for i in a)
	Max = max(abs(i) for i in a)
	L, R = -Max, Max
	eps = 10 ** -8
	for i in range(100):
	    m = (L + R) / 2
	    v1, v2 = f1(m), f2(m)
	    if abs(v1 - v2) < eps:
	        break
	    if v1 > v2:
	        L = m
	    else:
	        R = m
	
	print(v1)
main()
```