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;
double a[200001];
double weakness(double x) {
  double best_sum = 0.0;
  double curr_sum = 0.0;
  for (int i = 0; i < n; i++) {
    curr_sum += a[i] - x;
    best_sum = max(best_sum, curr_sum);
    if (curr_sum < 0.0) {
      curr_sum = 0.0;
    }
  }
  curr_sum = 0.0;
  for (int i = 0; i < n; i++) {
    curr_sum -= a[i] - x;
    best_sum = max(best_sum, curr_sum);
    if (curr_sum < 0.0) {
      curr_sum = 0.0;
    }
  }
  return best_sum;
}
double ternary_search(double l, double r) {
  double error = 2.0e-12;
  while (r - l > error) {
    double m1 = l + (r - l) / 3;
    double m2 = r - (r - l) / 3;
    double f1 = weakness(m1);
    double f2 = weakness(m2);
    if (f1 < f2)
      r = m2;
    else
      l = m1;
  }
  return weakness(l);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  t = 1;
  while (t--) {
    cin >> n;
    double mi = 2.0e6;
    double ma = -2.0e6;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      mi = min(a[i], mi);
      ma = max(a[i], ma);
    }
    double ans = ternary_search(mi - 1, ma + 1);
    ans *= 1.0e7;
    long long s = trunc(ans);
    ans = s / (1.0e7);
    std::cout << setprecision(7) << ans << endl;
  }
  return 0;
}