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;
void fio() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
double solve(double mid, vector<double> vec, double n) {
  for (int i = 0; i < n; i++) {
    vec[i] = vec[i] - mid;
  }
  double cur_sum = vec[0], max_sum = vec[0];
  double cur_min = vec[0], min_sum = vec[0];
  for (int i = 1; i < n; i++) {
    cur_sum = max(cur_sum + vec[i], vec[i]);
    max_sum = max(max_sum, cur_sum);
    cur_min = min(cur_min + vec[i], vec[i]);
    min_sum = min(cur_min, min_sum);
  }
  return max(max_sum, abs(min_sum));
}
int main() {
  fio();
  cout << fixed << setprecision(14);
  double n;
  cin >> n;
  vector<double> vec(n);
  for (int i = 0; i < n; i++) {
    cin >> vec[i];
  }
  double l = -(1e4 + 7), r = 1e4 + 7;
  for (int i = 0; i < 300; i++) {
    double m1 = l + (r - l) / 3;
    double m2 = r - (r - l) / 3;
    double f1 = solve(m1, vec, n);
    double f2 = solve(m2, vec, n);
    if (f1 < f2)
      r = m2;
    else
      l = m1;
  }
  double mid = r;
  for (int i = 0; i < n; i++) {
    vec[i] = vec[i] - mid;
  }
  double cur_sum = vec[0], max_sum = vec[0];
  double cur_min = vec[0], min_sum = vec[0];
  for (int i = 1; i < n; i++) {
    cur_sum = max(cur_sum + vec[i], vec[i]);
    max_sum = max(max_sum, cur_sum);
    cur_min = min(cur_min + vec[i], vec[i]);
    min_sum = min(cur_min, min_sum);
  }
  cout << max(max_sum, abs(min_sum)) << endl;
  return 0;
}