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 N = 2e5 + 11;
long long a[N];
int n;
double maxpoor(double x) {
  double min_sum = 0.0;
  double max_sum = 0.0;
  double cur_sum = 0.0;
  double ans = 0.0;
  for (int i = 1; i <= n; i++) {
    cur_sum += a[i] - x;
    ans = max(ans, fabs(cur_sum - min_sum));
    ans = max(ans, fabs(cur_sum - max_sum));
    min_sum = min(min_sum, cur_sum);
    max_sum = max(max_sum, cur_sum);
  }
  return ans;
}
int main() {
  while (1 == scanf("%d", &n)) {
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
    }
    double lo = -1e5, hi = 1e5;
    for (int i = 0; i < 200; i++) {
      double left = lo + (hi - lo) / 3.0;
      double right = hi - (hi - lo) / 3.0;
      if (maxpoor(left) < maxpoor(right))
        hi = right;
      else
        lo = left;
    }
    printf("%.8f\n", maxpoor(lo));
  }
  return 0;
}