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 N = 200010;
long double a[N];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n;
  long double ans;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  long double st = -10010, en = 10010;
  for (int t = 0; t < 70; t++) {
    long double m = (st + en) / 2;
    long double minn = 0, maxx = 0, sum = 0;
    for (int i = 0; i < n; i++) {
      sum += a[i] + m;
      minn = min(minn, sum);
      maxx = max(maxx, sum);
    }
    vector<int> v[2];
    if (minn == 0) v[0].push_back(-1);
    if (maxx == 0) v[1].push_back(-1);
    ans = maxx - minn;
    sum = 0;
    for (int i = 0; i < n; i++) {
      sum += a[i] + m;
      if (sum == minn) v[0].push_back(i);
      if (sum == maxx) v[1].push_back(i);
    }
    if (v[0].back() <= v[1][0]) en = m;
    if (v[1].back() <= v[0][0]) st = m;
  }
  cout << fixed << setprecision(10) << ans << endl;
  return 0;
}