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;
int A[400001];
double P[400001];
double f(double x) {
  P[0] = 0;
  double max_seg = 0, min_seg = 0, max_p = 0, min_p = 0;
  for (int i = 1; i <= n; i++) {
    P[i] = P[i - 1] + A[i] - x;
    max_p = max(max_p, P[i]);
    min_p = min(min_p, P[i]);
    max_seg = max(max_seg, P[i] - min_p);
    min_seg = min(min_seg, P[i] - max_p);
  }
  return max(max_seg, -min_seg);
}
double ternary_search(double l, double r, double eps) {
  double m1, m2, val1, val2;
  while (r - l >= eps) {
    m1 = l + (r - l) / 3;
    m2 = r - (r - l) / 3;
    val1 = f(m1);
    val2 = f(m2);
    if (val1 == val2) {
      l = m1;
      r = m2;
    } else if (val1 < val2) {
      r = m2;
    } else {
      l = m1;
    }
  }
  return l;
}
double ternary_search_max(double l, double r, double eps) {
  double m1, m2, val1, val2;
  while (r - l >= eps) {
    m1 = l + (r - l) / 3;
    m2 = r - (r - l) / 3;
    val1 = f(m1);
    val2 = f(m2);
    if (val1 == val2) {
      l = m1;
      r = m2;
    } else if (val1 < val2) {
      l = m1;
    } else {
      r = m2;
    }
  }
  return l;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  A[0] = 0;
  int maxi = 0;
  for (int i = 1; i <= n; i++) {
    cin >> A[i];
    if (A[i] > maxi) maxi = A[i];
  }
  cout.precision(13);
  cout << fixed << f(ternary_search(-10000, 10000, 1e-13)) << endl;
  return 0;
}