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;
double a[200000];
double b[200000];
double kadane() {
  double ans = b[0];
  double con = b[0];
  for (int i = 1; i < n; i++) {
    con = max(con + b[i], b[i]);
    ans = max(ans, con);
  }
  return ans;
}
double f(double x) {
  for (int i = 0; i < n; i++) {
    b[i] = a[i] - x;
  }
  double ans = abs(kadane());
  for (int i = 0; i < n; i++) {
    b[i] = -b[i];
  }
  ans = max(ans, abs(kadane()));
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  double L = -10000;
  double R = 10000;
  for (int i = 0; i < 100; i++) {
    double M1 = (2 * L + R) / 3;
    double M2 = (L + 2 * R) / 3;
    f(M1) < f(M2) ? R = M2 : L = M1;
  }
  cout << fixed << setprecision(7) << min(f(L), f(R)) << endl;
  return 0;
}