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 ar[200050];
double v[200050], ma[200050], mi[200050];
double fun(double x, int n) {
  int i;
  for (i = 0; i < n; i++) v[i] = ar[i] - x;
  for (i = 1; i < n; i++) v[i] += v[i - 1];
  double a = INT_MIN, b = INT_MAX;
  for (i = 0; i < n; i++) a = max(a, v[i]), b = min(b, v[i]);
  ma[0] = v[0], mi[0] = v[0];
  for (i = 0; i < n; i++)
    ma[i] = max(ma[i - 1], v[i]), mi[i] = min(mi[i - 1], v[i]);
  double c = INT_MIN, d = INT_MAX;
  for (i = n - 1; i >= 0; i--) {
    c = max(c, v[i]);
    a = max(a, c - mi[i]);
    d = min(d, v[i]);
    b = min(b, d - ma[i]);
  }
  a = max(abs(a), abs(b));
  return a;
}
double cal(int n) {
  int i;
  int p = 100000, q = -100000;
  for (i = 0; i < n; i++) p = min(p, ar[i]), q = max(q, ar[i]);
  double l = p, h = q;
  double x, y;
  for (i = 0; i < 71; i++) {
    x = 2 * l + h, y = l + 2 * h;
    x /= 3, y /= 3;
    if (fun(x, n) > fun(y, n))
      l = x;
    else
      h = y;
  }
  return fun(l, n);
}
int main() {
  int i, n;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d", &ar[i]);
  cout << setprecision(12) << fixed;
  cout << cal(n) << endl;
}