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, a[200010];
double calc(double val) {
  double minres = 200010 * 30010, maxres = -200010 * 30010, smin = 0, smax = 0,
         cur = 0;
  for (int i = 1; i <= n; ++i) {
    cur += a[i] - val;
    minres = min(minres, cur - smax);
    maxres = max(maxres, cur - smin);
    smin = min(cur, smin);
    smax = max(cur, smax);
  }
  return max(abs(minres), abs(maxres));
}
double ans() {
  double l = -20000.0, r = 20000.0;
  while (r - l > 1e-11) {
    double tl = (2 * l + r) / 3, tr = (2 * r + l) / 3;
    double res_tl = calc(tl), res_tr = calc(tr);
    if (res_tl - res_tr > 1e-11)
      l = tl;
    else if (res_tl - res_tr < 1e-11)
      r = tr;
    else if (res_tl - res_tr <= 1e-11)
      return res_tl;
  }
  return (l + r) / 2;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  printf("%0.6lf", calc(ans()));
}