Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 200010;
const double e = 1e-11;
double a[N];
int n;
double miin, maax;
double Min, Max;
void Do(double t) {
  Min = 1e9;
  Max = -1e9;
  double sum = 0;
  double in = 0, ax = 0;
  for (int i = 0; i < n; ++i) {
    sum += a[i] + t;
    in = min(in, sum);
    ax = max(ax, sum);
    Min = min(Min, sum - ax);
    Max = max(Max, sum - in);
  }
  if (!Min) Min = miin + t;
  if (!Max) Max = maax + t;
}
void binup() {
  double l = 0, r = 10000;
  while (r - l > e) {
    double m = (l + r) / 2;
    Do(m);
    if (-Min < Max)
      r = m;
    else
      l = m;
  }
  Do(l);
}
void bindown() {
  double l = -10000, r = 0;
  while (r - l > e) {
    double m = (l + r) / 2;
    Do(m);
    if (-Min > Max)
      l = m;
    else
      r = m;
  }
  Do(l);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  miin = 1e9;
  maax = -1e9;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    miin = min(miin, a[i]);
    maax = max(maax, a[i]);
  }
  Do(0);
  if (-Min < Max)
    bindown();
  else
    binup();
  printf("%0.12f", (Max - Min) / 2);
}