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;
vector<double> V;
double check(double x) {
  double cnt = 0;
  double mins = 0, maxs = 0, ans = 0;
  for (int i = 0; i < V.size(); i++) {
    cnt += V[i] * 1.00 - x;
    ans = max(fabs(cnt - maxs), ans);
    ans = max(fabs(cnt - mins), ans);
    mins = min(cnt, mins);
    maxs = max(cnt, maxs);
  }
  return ans;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    double x;
    cin >> x;
    x *= 1.00;
    V.push_back(x);
  }
  double le = -1e6, ri = 1e6, res = -1;
  for (int i = 0; i < 250; i++) {
    double l1 = le + (le - ri) / 3.00;
    double l2 = le + 2 * (le - ri) / 3.00;
    if (abs(check(l1)) < abs(check((l2))))
      ri = l2;
    else
      le = l1;
  }
  res = check((le + ri) / 2);
  printf("%.10lf\n", abs(res));
  return 0;
}