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;
double f(double m, vector<int> &v) {
  int n = v.size();
  vector<long double> v2(n);
  for (int i = 0; i < (n); i++) v2[i] = v[i] - m;
  long double mx_otr = 0;
  long double mn_otr = 0;
  long double cur = 0;
  for (int i = 0; i < (n); i++) {
    cur += v2[i];
    if (cur < 0) cur = 0;
    mx_otr = max(mx_otr, cur);
  }
  cur = 0;
  for (int i = 0; i < (n); i++) {
    cur += v2[i];
    if (cur >= 0) cur = 0;
    mn_otr = min(mn_otr, cur);
  }
  return max(mx_otr, abs(mn_otr));
}
inline void solve() {
  int n;
  cin >> n;
  vector<int> v(n);
  for (int i = 0; i < (n); i++) cin >> v[i];
  long double l = -10000, r = 10000;
  while (abs(r - l) > 1e-12) {
    long double m1 = l + (r - l) / 3;
    long double m2 = r - (r - l) / 3;
    if (f(m1, v) <= f(m2, v))
      r = m2;
    else
      l = m1;
  }
  cout << fixed << setprecision(10) << f(l + (r - l) / 2.0, v);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int q = 1;
  while (q--) {
    solve();
    cout << '\n';
  }
}