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>
struct compMinPQ {
  bool operator()(const int &a, const int &b) { return a > b; }
};
using namespace std;
double a[200005];
int N;
double f(double x) {
  int i, j, k;
  for (i = 1; i <= N; i++) a[i] = a[i] - x;
  double max_so_far = a[1], max_ending = a[1];
  for (i = 2; i <= N; i++) {
    max_ending = max(a[i], max_ending + a[i]);
    max_so_far = max(max_so_far, max_ending);
  }
  double min_so_far = a[1], min_ending = a[1];
  for (i = 2; i <= N; i++) {
    min_ending = min(a[i], min_ending + a[i]);
    min_so_far = min(min_ending, min_so_far);
  }
  for (i = 1; i <= N; i++) a[i] = a[i] + x;
  return max(abs(max_so_far), abs(min_so_far));
}
int main() {
  int i, j, k;
  cin >> N;
  for (i = 1; i <= N; i++) cin >> a[i];
  double low = -10005;
  double high = 10005;
  double esp = 1e-13;
  double ans;
  for (i = 0; i < 200; i++) {
    double mid1 = low + (high - low) / 3.0;
    double mid2 = high - (high - low) / 3.0;
    if (f(mid1) - f(mid2) > esp) {
      ans = mid2;
      low = mid1 + esp;
    } else {
      ans = mid1;
      high = mid2 - esp;
    }
  }
  ans = f(ans);
  printf("%.12f\n", ans);
  return 0;
}