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;
long long n;
vector<double> a, p;
double poor(double x) {
  p[1] = a[1] - x;
  for (long long i = 2; i <= n; i++) p[i] = p[i - 1] + a[i] - x;
  long long m1 = 0;
  long long m2 = 0;
  double mn = p[1];
  double mx = p[1];
  for (long long i = 1; i <= n; i++) {
    mx = max(mx, p[i] - p[m1]);
    mn = min(mn, p[i] - p[m2]);
    if (p[i] < p[m1]) m1 = i;
    if (p[i] > p[m2]) m2 = i;
  }
  return max(abs(mx), abs(mn));
}
signed main() {
  cin >> n;
  a.resize(n + 1);
  p.resize(n + 1);
  for (long long i = 1; i <= n; i++) cin >> a[i];
  double l = -10000, r = 10000;
  for (long long i = 0; i < 300; i++) {
    double left = l + (r - l) / 3;
    double right = l + 2 * (r - l) / 3;
    if (poor(left) < poor(right))
      r = right;
    else
      l = left;
  }
  cout << fixed << setprecision(10) << poor(l);
  return 0;
}