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.
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 100, oo = 1e9;
long double a[N], b[N];
int n;
inline long double calc(const long double &x) {
  long double min_pref = 0, mx_pref = 0;
  long double sum = 0;
  long double ans = 0;
  for (int i = 1; i <= n; ++i) {
    sum += a[i] - x;
    min_pref = min(min_pref, sum);
    mx_pref = max(mx_pref, sum);
    ans = max(ans, max(fabs(sum - min_pref), fabs(sum - mx_pref)));
  }
  return ans;
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  long double l = -1e5, r = 1e5;
  for (int cnt = 0; cnt <= 200; ++cnt) {
    long double mid1 = l + (r - l) / 3;
    long double mid2 = r - (r - l) / 3;
    if (calc(mid1) < calc(mid2)) {
      r = mid2;
    } else {
      l = mid1;
    }
  }
  cout.precision(10);
  cout << fixed;
  cout << calc(l);
  return 0;
}
