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 double EPS = 1e-9;
double g[200100], f[200100], a[233333];
int n;
double equ(double x) {
  double ans = 0;
  g[0] = 0;
  f[0] = 0;
  double maxn = -1e20, minn = 1e20;
  for (int i = 1; i <= n; i++) {
    f[i] = max(f[i - 1] + a[i] - x, a[i] - x);
    maxn = max(maxn, f[i]);
    g[i] = min(g[i - 1] + a[i] - x, a[i] - x);
    minn = min(minn, g[i]);
  }
  ans = max(maxn, fabs(minn));
  return ans;
}
double ternarySearch(double l, double r) {
  for (int i = 1; i <= 233; i++) {
    double ll = (2 * l + r) / 3;
    double rr = (l + 2 * r) / 3;
    double ans1 = equ(ll);
    double ans2 = equ(rr);
    if (ans1 > ans2)
      l = ll;
    else
      r = rr;
  }
  return l;
}
int main() {
  scanf("%d", &n);
  double maxn = 100000.0;
  for (int i = 1; i <= n; i++) {
    scanf("%lf", &a[i]);
  }
  double ans = ternarySearch(-maxn, maxn);
  ans = equ(ans);
  printf("%.8lf\n", ans);
  return 0;
}