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;
const double pi = acos(-1.0);
const double eps = 1e-11;
double calc(vector<double> &v, double nm) {
  int n = int((v).size());
  double mx = 0;
  double cur = 0;
  for (int i = 0; i < n; i++) {
    if (cur < 0) {
      cur = 0;
    }
    cur += (v[i] - nm);
    mx = max(mx, cur);
  }
  cur = 0;
  for (int i = 0; i < n; i++) {
    if (cur > 0) {
      cur = 0;
    }
    cur += (v[i] - nm);
    mx = max(mx, abs(cur));
  }
  return mx;
}
int main() {
  std::ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  vector<double> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  double l = -1000000000, r = 1000000000;
  for (int i = 0; i < 350; i++) {
    double m1 = l + (r - l) / 3, m2 = r - (r - l) / 3;
    if (calc(v, m1) >= calc(v, m2)) {
      l = m1;
    } else
      r = m2;
  }
  cout << fixed;
  cout << setprecision(14);
  cout << min(calc(v, l), calc(v, r)) << endl;
  return 0;
}