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 double eps = 1e-11;
vector<double> a;
double okk(double m) {
  long long n = a.size();
  double ans = 0;
  double tmp[n], pf[n];
  for (long long i = 0; i < n; i++) {
    tmp[i] = a[i] - m;
    if (i == 0)
      pf[i] = tmp[i];
    else
      pf[i] = pf[i - 1] + tmp[i];
  }
  double mn = 0;
  for (long long i = 0; i < n; i++) {
    ans = max(ans, pf[i] - mn);
    mn = min(mn, pf[i]);
    pf[i] *= -1;
  }
  mn = 0;
  for (long long i = 0; i < n; i++) {
    ans = max(ans, pf[i] - mn);
    mn = min(mn, pf[i]);
  }
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << setprecision(15) << fixed;
  long long n;
  cin >> n;
  a.resize(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  double l = -1e6, r = 1e6, ans = 1e15;
  while (r - l > eps) {
    double m1 = l + (r - l) / 3;
    double m2 = r - (r - l) / 3;
    double t1 = okk(m1);
    double t2 = okk(m2);
    ans = min(ans, t1);
    ans = min(ans, t2);
    if (t1 > t2)
      l = m1;
    else
      r = m2;
  }
  cout << ans << '\n';
}