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, 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.

The poorness:连续子序列的和的绝对值。

题解：ternary searc（三分查找），很典型的三分题。x取极值时最小，否则都会增大，所以是一个凹函数。

1 #include
       
         2 using namespace std; 3  4 const int maxn=2e5+5; 5  6 int n; 7 int a[maxn]; 8  9 double maxSum(double A[]){10     double ans=0,tem=0;11     for(int i=1;i<=n;i++){12         tem+=A[i];13         if(tem<0) tem=0;14         ans=max(ans,tem);15     }16     return ans;17 }18 19 double compute(double x){20     double b[maxn];21     for(int i=1;i<=n;i++) b[i]=a[i]-x;22     double ans1=maxSum(b);23     for(int i=1;i<=n;i++) b[i]=-b[i];24     double ans2=maxSum(b);25     return max(ans1,ans2);26 }27 28 int main()29 {   scanf("%d",&n);30     for(int i=1;i<=n;i++) scanf("%d",&a[i]);31     double l=-1e4,r=1e4;32     for(int i=1;i<=100;i++){33         double midl=(2*l+r)/3.0;34         double midr=(2*r+l)/3.0;35         if(compute(midl)