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 arr[200005];
int n;
double solve(double x) {
  double ans1 = 0, ans2 = 0;
  double val = 0;
  for (int i = 0; i < n; i++) {
    val += (arr[i] - x);
    if (val > ans1) ans1 = val;
    if (val < 0) val = 0;
  }
  val = 0;
  for (int i = 0; i < n; i++) {
    val += (arr[i] - x);
    if (val < ans2) ans2 = val;
    if (val > 0) val = 0;
  }
  return max(ans1, -ans2);
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  double l1 = -100000, l2 = 100000;
  double mid1, mid2;
  for (int i = 0; i < 100; i++) {
    mid1 = (l1 + l2) / 2.0;
    mid2 = (mid1 + l2) / 2.0;
    if (solve(mid1) < solve(mid2)) {
      l2 = mid2;
    } else {
      l1 = mid1;
    }
  }
  cout << fixed << setprecision(9) << solve(l2) << endl;
  return 0;
}