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 X = 100000 + 5;
long double foo1(vector<long double> &v) {
  long double s = 0, mx = 0;
  for (int i = 0; i < (int)v.size(); i++) {
    s += v[i];
    mx = max(mx, s);
    if (s < 1e-10) s = 0;
  }
  return mx;
}
long double foo(long double x, vector<long double> v) {
  for (int i = 0; i < (int)v.size(); i++) v[i] = v[i] - x;
  long double s = foo1(v);
  for (int i = 0; i < (int)v.size(); i++) v[i] *= -1;
  s = max(s, foo1(v));
  return s;
}
long double calc(vector<long double> &v) {
  long double lo = -1e5, hi = 1e5;
  for (int it = 1; it < 100; it++) {
    long double ml = lo + (hi - lo) / 3.0, mr = hi - (hi - lo) / 3.0;
    if (foo(ml, v) <= foo(mr, v))
      hi = mr;
    else
      lo = ml;
  }
  return foo(lo, v);
}
void MAIN() {
  int n;
  long double x;
  cin >> n;
  vector<long double> v;
  for (int i = 0; i < n; i++) {
    cin >> x;
    v.push_back(x);
  }
  cout << calc(v);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << setprecision(12);
  MAIN();
  return 0;
}