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 eps = 1e-9;
long long int n;
long double arr[1000000];
long double solve(long double mid) {
  double maxi = 0.0, sum = 0.0;
  for (long long int i = 0; i < n; i++) {
    sum += (arr[i] - mid);
    if (sum < 0) {
      sum = 0;
    }
    maxi = max(abs(sum), maxi);
  }
  sum = 0.0;
  for (long long int i = 0; i < n; i++) {
    sum += (arr[i] - mid);
    if (sum > 0) {
      sum = 0;
    }
    maxi = max(abs(sum), maxi);
  }
  return maxi;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  long double lo = -10000.0, hi = 10000.0, mid1, mid2;
  for (long long int i = 1; i <= 100; i++) {
    mid1 = lo + (hi - lo) / 3;
    mid2 = hi - (hi - lo) / 3;
    if (solve(mid1) >= solve(mid2)) {
      lo = mid1;
    } else {
      hi = mid2;
    }
  }
  cout << fixed << setprecision(10) << solve(lo) << endl;
  return 0;
}