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 i, n, a[200005];
double st = -1e5, dr = 1e5;
double ABS(double a) { return a > 0 ? a : -a; }
double Check1(double x) {
  double aux = 0, now = 0;
  for (int i = 1; i <= n; ++i) {
    now += a[i] - x;
    aux = max(aux, now);
    if (now < 0) now = 0;
  }
  return aux;
}
double Check2(double x) {
  double aux = 0, now = 0;
  for (int i = 1; i <= n; ++i) {
    now += x - a[i];
    aux = max(aux, now);
    if (now < 0) now = 0;
  }
  return aux;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (i = 1; i <= n; ++i) cin >> a[i];
  while (st <= dr) {
    double pivot = (st + dr) / 2.0;
    if (Check1(pivot) > Check2(pivot))
      st = pivot + 1e-12;
    else
      dr = pivot - 1e-12;
  }
  cout << setprecision(7) << fixed
       << max(Check1((st + dr) / 2.0), Check2((st + dr) / 2.0)) << '\n';
  return 0;
}
