Weakness and Poorness CodeForces

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）https://www.cnblogs.com/whywhy/p/4886641.html，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<bits/stdc++.h>
 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)<compute(midr)) r=midr;
36         else l=midl;
37     }
38     printf("%.7f
",compute(r));
39 }