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;
double find_max(int *a, int n, double x) {
  double min_p = a[0] - x;
  double max_p = a[0] - x;
  double maxv = abs(a[0] - x);
  for (int i = 1; i < n; ++i) {
    min_p = min(min_p + a[i] - x, a[i] - x);
    max_p = max(max_p + a[i] - x, a[i] - x);
    maxv = max(abs(min_p), maxv);
    maxv = max(abs(max_p), maxv);
  }
  return maxv;
}
int main() {
  ios_base::sync_with_stdio(false);
  int a[200000];
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> a[i];
  double L = -1e9;
  double R = 1e9 + 1e-2;
  while (abs(L - R) > 1e-11) {
    double x1 = L + (R - L) * 1 / 3;
    double x2 = L + (R - L) * 2 / 3;
    double vl = find_max(a, n, x1);
    double vr = find_max(a, n, x2);
    if (vl + 1e-9 < vr) {
      R = x2;
    } else {
      L = x1;
    }
  }
  std::cout.precision(8);
  cout << fixed << find_max(a, n, L) << endl;
  return 0;
}