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 t[200005];
int n, k;
double cal(double x) {
  double maxx = 0, minn = 0;
  for (int i = 1; i <= n; i++) {
    double val = -x * i + t[i];
    maxx = max(maxx, val);
    minn = min(minn, val);
  }
  return maxx - minn;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  t[0] = 0;
  for (int i = 1; i <= n; i++) cin >> k, t[i] = t[i - 1] + k;
  double l = -30000, h = 30000;
  int cc = 0;
  for (int i = 1; i <= 100; i++) {
    cc++;
    double r1 = l + (h - l) / 3;
    double r2 = l + (h - l) * 2 / 3;
    double R1 = cal(r1);
    double R2 = cal(r2);
    if (R1 < R2 + 1e-8)
      h = r2;
    else
      l = r1;
  }
  printf("%.9lf", cal((l + h) / 2));
  return 0;
}