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 int N = 211111;
int a[N];
double weakness(double x, int n) {
  double sum = 0;
  double last_min = 0;
  double last_max = 0;
  double ans_min = 0;
  double ans_max = 0;
  for (int i = 0; i < n; i++) {
    sum += a[i] - x;
    ans_max = max(ans_max, sum - last_min);
    ans_min = min(ans_min, sum - last_max);
    last_min = min(last_min, sum);
    last_max = max(last_max, sum);
  }
  return max(ans_max, -ans_min);
}
int main() {
  ios ::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  double left = -20000.0;
  double right = 20000.0;
  for (int t = 0; t < 200; t++) {
    double mid1 = (left * 2 + right) / 3;
    double mid2 = (left + right * 2) / 3;
    double w1 = weakness(mid1, n);
    double w2 = weakness(mid2, n);
    if (w1 > w2)
      left = mid1;
    else
      right = mid2;
  }
  cout << setprecision(12) << weakness(left, n) << '\n';
  return 0;
}