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 <typename T1, typename T2>
std::istream& operator>>(std::istream& in, pair<T1, T2>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T1, typename T2>
std::ostream& operator<<(std::ostream& out, pair<T1, T2>& a) {
  out << a.first << " " << a.second;
  return out;
}
const long long N = 2e5 + 5;
long long a[N];
double b[N];
long long n;
double calc(double x) {
  for (long long i = 0; i < n; i++) b[i + 1] = a[i] - x;
  double s = 0, ans = 0;
  b[0] = 0;
  for (long long i = 1; i < n + 1; i++) {
    b[i] += b[i - 1];
  }
  double l = 0, u = 0;
  for (long long i = 1; i < n + 1; i++) {
    ans = max(ans, abs(b[i] - l));
    ans = max(ans, abs(b[i] - u));
    l = min(l, b[i]);
    u = max(u, b[i]);
  }
  return ans;
}
void solve() {
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> a[i];
  double L = -10002, U = 10002;
  for (long long i = 0; i < 90; i++) {
    double d = (U - L) / 3;
    if (calc(L + d) > calc(L + 2 * d)) {
      L = L + d;
    } else {
      U = U - d;
    }
  }
  cout << calc(L);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(9);
  cout << fixed;
  long long t = 1;
  while (t--) solve();
  return 0;
}