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;
template <class T>
bool uin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
double f(const vector<int> &a, double x) {
  double s = 0.0, mins = 0, maxs = 0;
  for (int i = 0; i < (int)(a.size()); ++i) {
    s += a[i] + x;
    uin(mins, s);
    uax(maxs, s);
  }
  return maxs - mins;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.precision(10);
  cout << fixed;
  int N;
  cin >> N;
  vector<int> a(N);
  for (int i = 0; i < (int)(N); ++i) cin >> a[i];
  double l = -1e9, r = 1e9;
  int A = 11, B = 10;
  for (int its = 0; its < (int)(100); ++its) {
    double m1 = (A * l + B * r) / (A + B);
    double m2 = (B * l + A * r) / (A + B);
    double f1 = f(a, m1), f2 = f(a, m2);
    if (f1 < f2)
      r = m2;
    else
      l = m1;
  }
  cout << f(a, l) << '\n';
  return 0;
}