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 MAX = 2e5 + 9;
int n;
double a[MAX], sum[MAX];
double cal(double X) {
  double hsgs = 0, Mn = 0, Mx = 0;
  for (int i = 1; i <= n; i++) {
    sum[i] = sum[i - 1] + a[i] - X;
    hsgs = max(max(fabs(sum[i] - Mn), fabs(sum[i] - Mx)), hsgs);
    Mn = min(Mn, sum[i]), Mx = max(Mx, sum[i]);
  }
  return hsgs;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  double l = -1e9 - 9, r = 1e9 + 9;
  for (int i = 1; i <= 250; i++)
    if (cal((2 * l + r) / 3) < cal((l + 2 * r) / 3))
      r = (l + 2 * r) / 3;
    else
      l = (2 * l + r) / 3;
  cout << fixed << setprecision(10) << cal(l);
}