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 INF = 1e9;
vector<int> a;
int n;
double Solve(double x) {
  double _max = 0.0;
  double _temp = 0.0;
  double sum = 0.0;
  for (int i = 0; i < n; i++) {
    sum = max(sum + a[i] - x, a[i] - x);
    _max = max(_max, sum);
  }
  sum = 0.0;
  for (int i = 0; i < n; i++) {
    sum = max(sum + x - a[i], x - a[i]);
    _temp = max(_temp, sum);
  }
  return max(_max, _temp);
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    a.emplace_back(x);
  }
  double left = -INF;
  double right = INF;
  double mid_L, mid_R;
  double fx, fy;
  for (int i = 0; i < 300; i++) {
    mid_L = left + (right - left) / 3;
    mid_R = right - (right - left) / 3;
    fx = Solve(mid_L);
    fy = Solve(mid_R);
    if (fx < fy)
      right = mid_R;
    else
      left = mid_L;
  }
  cout << setprecision(11) << fixed << fx << '\n';
  return 0;
}