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 c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
long double check(long double val, vector<long double>& a) {
  int n = a.size();
  for (int i = 0; i < n; i++) {
    a[i] -= val;
  }
  long double mx = -1e9;
  long double curr = -1e9;
  long double mn = 1e9;
  for (int i = 0; i < n; i++) {
    curr = max(curr + a[i], a[i]);
    mx = max(curr, mx);
  }
  curr = 1e9;
  for (int i = 0; i < n; i++) {
    curr = min(curr + a[i], a[i]);
    mn = min(curr, mn);
  }
  for (int i = 0; i < n; i++) {
    a[i] += val;
  }
  return max(abs(mn), abs(mx));
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<long double> a(n);
  for (int i = 0; i < n; i++) {
    int chi;
    cin >> chi;
    a[i] = (long double)chi;
  }
  long double l = -100005.00, r = 100005.00;
  for (int i = 0; i < 350; i++) {
    long double m1 = l + (r - l) / 3.0;
    long double m2 = r - (r - l) / 3.0;
    long double f1 = check(m1, a);
    long double f2 = check(m2, a);
    if (f1 > f2) {
      l = m1;
    } else
      r = m2;
  }
  cout << fixed << setprecision(9) << check(r, a) << endl;
}