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;
int n;
int arr[200000];
long double sum(long double x) {
  long double maxF = 0, maxE = 0;
  long double minF = 0, minE = 0;
  for (int i = int(0); i < int(n); i++) {
    maxE = max(arr[i] - x, maxE + arr[i] - x);
    maxF = max(maxF, maxE);
    minE = min(arr[i] - x, minE + arr[i] - x);
    minF = min(minF, minE);
  }
  return max(abs(maxF), abs(minF));
}
long double binSearch() {
  long double lo = -10000, hi = 10000;
  long double x1, x2, s1, s2;
  while (abs(hi - lo) > 1e-12) {
    x1 = (lo + lo + hi) / 3;
    x2 = (lo + hi + hi) / 3;
    s1 = sum(x1);
    s2 = sum(x2);
    if (s1 < s2)
      hi = x2;
    else
      lo = x1;
  }
  return s1;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cin.tie(0), cout.tie(0),
      cout.precision(7);
  while (cin >> n) {
    for (int i = int(0); i < int(n); i++) cin >> arr[i];
    cout << binSearch() << endl;
  }
  return 0;
}