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;
const long long mod = 1e9 + 7;
const double oo = 1e5 + 1;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<double> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  auto f = [&](double x) {
    double Max = 0, Min = 0, cur = 0, ans = 0;
    for (long long i = 0; i < n; i++) {
      cur += a[i] - x;
      ans = max(ans, abs(cur - Min));
      ans = max(ans, abs(cur - Max));
      Min = min(Min, cur);
      Max = max(Max, cur);
    }
    return ans;
  };
  double l = -1e4, r = 1e4;
  for (long long i = 1; i <= 300; i++) {
    double mid1 = l + (r - l) / 3;
    double mid2 = r - (r - l) / 3;
    if (f(mid1) < f(mid2))
      r = mid2;
    else
      l = mid1;
  }
  cout << fixed << setprecision(10) << f(l);
}