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 int MAXN = 2e5 + 5;
double a[MAXN], b[MAXN];
int n;
void input();
double f(double);
void solve();
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  input();
  solve();
  return 0;
}
void input() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
}
double f(double x) {
  double res, mn, mx;
  res = abs(a[0] - x);
  b[0] = mn = mx = a[0] - x;
  mx = max(0.0, mx);
  mn = min(0.0, mn);
  for (int i = 1; i < n; i++) {
    b[i] = b[i - 1] + a[i] - x;
    res = max(res, max(abs(b[i] - mn), abs(mx - b[i])));
    mn = min(mn, b[i]);
    mx = max(mx, b[i]);
  }
  return res;
}
void solve() {
  double l = -1e4 - 1, r = 1e4 + 1, mid = (l + r) / 2, eps = 1e-12;
  for (int i = 0; i < 100; i++) {
    (f(mid) < f(mid + eps)) ? r = mid : l = mid;
    mid = (l + r) / 2;
  }
  cout.precision(15);
  cout << fixed << f(l);
}