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;
template <class P, class Q>
inline void smin(P &a, Q b) {
  if (b < a) a = b;
}
template <class P, class Q>
inline void smax(P &a, Q b) {
  if (a < b) a = b;
}
const int maxn = 200000 + 100;
int n;
int a[maxn], s[maxn];
double f(double x) {
  double mn = 0, mx = 0;
  for (int i = 0, _n = (int)(n + 1); i < _n; i++)
    smin(mn, s[i] - i * x), smax(mx, s[i] - i * x);
  return mx - mn;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0, _n = (int)(n); i < _n; i++) cin >> a[i];
  for (int i = 0, _n = (int)(n); i < _n; i++) s[i + 1] = s[i] + a[i];
  double lo = -1e5, hi = 1e5;
  for (int z = 0, _n = (int)(100); z < _n; z++) {
    double m1 = (2 * lo + hi) / 3;
    double m2 = (lo + 2 * hi) / 3;
    if (f(m1) < f(m2))
      hi = m2;
    else
      lo = m1;
  }
  cout.setf(ios::fixed);
  cout.precision(9);
  cout << f(lo) << '\n';
  return 0;
}