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;
int arr[200001];
long double cum[200001];
long double EPS = 1e-12;
long double f(long double mid) {
  cum[0] = arr[0] - mid;
  long double mini = 0;
  long double maxi = 0;
  long double ans = 0;
  for (int i = 0; i < n; i++) {
    if (i != 0) cum[i] = (arr[i] - mid) + cum[i - 1];
    ans = max(ans, abs(cum[i] - maxi));
    ans = max(ans, abs(cum[i] - mini));
    maxi = max(maxi, cum[i]);
    mini = min(mini, cum[i]);
  }
  return ans;
}
long double ternarySearch(long double left, long double right) {
  while (right - left > EPS) {
    long double g = left + (right - left) / 3,
                h = left + 2 * (right - left) / 3;
    if (f(g) < f(h))
      right = h;
    else
      left = g;
  }
  return f((left + right) / 2);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> arr[i];
  cout << fixed << setprecision(10) << ternarySearch(-10000, 10000) << endl;
  return 0;
}