Weakness and Poorness CodeForces - 578C 三分搜索 （精度!)

You are given a sequence of n integers a₁, a₂, ..., a_n.

Determine a real number x such that the weakness of the sequence a₁ - x, a₂ - x, ..., a_n - 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, ..., a_n (|a_i| ≤ 10 000).

Output

Output a real number denoting the minimum possible weakness of a₁ - x, a₂ - x, ..., a_n - 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 = 0; i < n; i++)
        tmp[i] = a[i] - x;
    double cur = 0, ans = 0;
    for (int i = 0; i < n; i++)
    {
        cur = cur + tmp[i];
        if (cur < 0)
            cur = 0;
        ans = max(cur, ans);
    }
    cur = 0;
    for (int i = 0; i < n; i++)
    {
        cur = cur - tmp[i];
        if (cur < 0)
            cur = 0;
        ans = max(cur, ans);
    }
    return ans;
}
int main()
{
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
        scanf("%lf", &a[i]);
    double beg = -10005, end = 10005;
    int time = 100;
    while (time--)
    {
        double ml = (beg + beg + end) / 3, mr = (end + end + beg) / 3;
        if (cal(ml) > cal(mr))
            beg = ml;
        else
            end = mr;
    }
    printf("%.15lf\n", cal(beg));
    return 0;
}