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;
int n;
long double a[200005], b[200005];
long double fun(long double x) {
  for (int i = 0; i < n; i++) {
    b[i] = a[i] - x;
  }
  long double msum = b[0];
  long double ans = b[0];
  int i;
  for (i = 1; i < n; i++) {
    msum = max(b[i], msum + b[i]);
    ans = max(ans, msum);
  }
  long double msum1 = b[0];
  long double ans1 = b[0];
  for (i = 1; i < n; i++) {
    msum1 = min(msum1 + b[i], b[i]);
    ans1 = min(ans1, msum1);
  }
  return max(abs(ans), abs(ans1));
}
int main() {
  cin >> n;
  int i;
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  long double lo = -2000000200, hi = 2000000200;
  long double m1, m2, ans;
  long double eps = 1e-13;
  while (hi - lo > eps) {
    m1 = lo + (hi - lo) / 3;
    m2 = hi - (hi - lo) / 3;
    long double val1 = fun(m1);
    long double val2 = fun(m2);
    if (val1 < val2) {
      hi = m2;
    } else {
      lo = m1;
    }
  }
  ans = fun(lo);
  cout << fixed << setprecision(10) << ans;
  return 0;
}