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.

Example

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, 1and 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.

#include<cstdio>
#include<algorithm>
#include<vector>
#include<iostream>
#define MAXN 200009
#define eps 1e-11 + 1e-12/2
typedef long long LL; using namespace std;
/*
要求a[i]减去某个数字x后的最大字段和的最小绝对值!
s(a[i]-x)是单调的,加上绝对值之后变成单谷函数,三分搜索
*/ int n;
double a[MAXN],tmp[MAXN];
double cal(double x)
{
for (int i = ; i < n; i++)
tmp[i] = a[i] - x;
double cur = , ans = ;
for (int i = ; i < n; i++)
{
cur = cur + tmp[i];
if (cur < )
cur = ;
ans = max(cur, ans);
}
cur = ;
for (int i = ; i < n; i++)
{
cur = cur - tmp[i];
if (cur < )
cur = ;
ans = max(cur, ans);
}
return ans;
}
int main()
{
scanf("%d", &n);
for (int i = ; i < n; i++)
scanf("%lf", &a[i]);
double beg = -, end = ;
int time = ;
while (time--)
{
double ml = (beg + beg + end) / , mr = (end + end + beg) / ;
if (cal(ml) > cal(mr))
beg = ml;
else
end = mr;
}
printf("%.15lf\n", cal(beg));
return ;
}

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