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[200005], sp[200005];
double val(double x) {
  double mx = sp[0], mn = sp[0];
  for (int i = 1; i <= n; i++) {
    mx = (((mx) > (sp[i] - i * x)) ? (mx) : (sp[i] - i * x));
    mn = (((mn) < (sp[i] - i * x)) ? (mn) : (sp[i] - i * x));
  }
  return mx - mn;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    sp[i] = sp[i - 1] + a[i];
  }
  double l = -10002, r = 10002, mid1, mid2;
  for (int iter = 1; iter <= 200; iter++) {
    double x = (r - l) / 3;
    mid1 = l + x;
    mid2 = r - x;
    double xx = val(mid1), yy = val(mid2);
    if (fabs(xx - yy) < 1e-9) break;
    if (xx < yy)
      r = mid2;
    else
      l = mid1;
  }
  printf("%.10f\n", val((l + r) / 2));
  return 0;
}