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.
#include <bits/stdc++.h>
using namespace std;
int a[200005];
int n;
const double eps = 1e-12;
double fun(double x) {
  double tmp = 0, ans = 0, tmin = 0, tmax = 0;
  for (int i = 0; i < n; ++i) {
    tmp += (a[i] - x);
    ans = max(ans, fabs(tmp - tmin));
    ans = max(ans, fabs(tmp - tmax));
    tmin = min(tmin, tmp);
    tmax = max(tmax, tmp);
  }
  return ans;
}
int main() {
  double l, r;
  scanf("%d", &n);
  l = 1e7;
  r = -1e7;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    l = min(l, 1. * a[i]);
    r = max(r, 1. * a[i]);
  }
  double f1, f2, mid1, mid2;
  for (int i = 0; i < 500; ++i) {
    mid1 = l + (r - l) / 3;
    mid2 = r - (r - l) / 3;
    f1 = fun(mid1);
    f2 = fun(mid2);
    if (f1 < f2)
      r = mid2;
    else
      l = mid1;
  }
  cout.precision(10);
  cout << fixed << fun(l) << endl;
  return 0;
}
