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 a[500008];
long long int n, m, k;
double b[500008];
double f(double mid) {
  double sum = 0.0;
  double arr[n + 7];
  for (int i = 1; i <= n; i++) {
    arr[i] = a[i] - mid;
  }
  double mn = 0;
  double maax = 0;
  for (int i = 1; i <= n; i++) {
    mn += arr[i];
    maax += arr[i];
    mn = min(mn, arr[i]);
    maax = max(maax, arr[i]);
    sum = max(sum, abs(mn));
    sum = max(sum, abs(maax));
  }
  return sum;
}
int main() {
  long long int t = 1;
  while (t--) {
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];
    double lo = -1e9;
    double hi = 1e9;
    double midl, midr;
    double ans = 1234567891234;
    for (int i = 1; i <= 200; i++) {
      midl = lo + (hi - lo) / 3;
      midr = hi - (hi - lo) / 3;
      if (f(midl) > f(midr)) {
        lo = midl;
      } else
        hi = midr;
      ans = min(ans, f(midl));
      ans = min(ans, f(midr));
    }
    cout << setprecision(15) << fixed;
    cout << ans << endl;
  }
}