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 N = 5e6 + 1000;
long long a[N], n;
long double f(long double x) {
  long double ans = 0;
  long double maxi = 0, mini = 0;
  long double sum = 0;
  for (long long i = 0; i < n; i++) {
    sum += a[i] + x;
    ans = max(ans, max(abs(sum - maxi), abs(sum - mini)));
    mini = min(mini, sum);
    maxi = max(maxi, sum);
  }
  return ans;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  long double r = 2e9;
  long double l = -r;
  for (long long _ = 0; _ < 200; _++) {
    long double l2 = (2 * l + r) / 3;
    long double r2 = (l + 2 * r) / 3;
    if (f(l2) <= f(r2))
      r = r2;
    else
      l = l2;
  }
  cout << setprecision(20) << f(l) << "\n";
  return 0;
}