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 maxn = 200000 + 24;
int n;
double a[maxn];
double cal(double x) {
  double ans1 = 0.0, tmp1 = 0.0, ans2 = 0.0, tmp2 = 0.0;
  for (int i = 1; i <= n; ++i) {
    double k = a[i] - x;
    tmp1 = max(tmp1 + k, 0.0);
    tmp2 = max(tmp2 - k, 0.0);
    ans1 = max(ans1, tmp1);
    ans2 = max(ans2, tmp2);
  }
  return max(ans1, ans2);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%lf", &a[i]);
  double eps = 1e-6 / n / 2 * 1.0;
  double l = -10000.0, r = 10000.0, m1 = 0.0, m2 = 0.0, ans = 0.0;
  while (r - l > eps) {
    m1 = l + (r - l) / 3.0;
    m2 = r - (r - l) / 3.0;
    if (cal(m1) >= cal(m2))
      l = ans = m1;
    else
      r = ans = m2;
  }
  printf("%.10lf\n", cal(ans));
}