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 + 10;
int a[N], n;
double solve(double x) {
  double ma = -1, now = 0;
  for (int i = 0; i < n; i++) ma = max(ma, now = max(0.0, now + a[i] - x));
  now = 0;
  for (int i = 0; i < n; i++) ma = max(ma, now = max(0.0, now + x - a[i]));
  return ma;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  double l = -1e4, r = 1e4;
  int T = 200;
  while (T--) {
    double m = (l + r) / 2, ml = (l + m) / 2, mr = (r + m) / 2;
    if (solve(ml) <= solve(mr))
      r = mr;
    else
      l = ml;
  }
  printf("%.15f\n", solve(l));
  return 0;
}