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;
int n;
double a[200005], b[200005];
double l = -10001, r = 10001;
const double eps = 2e-12;
double solve(double x) {
  for (int i = 1; i <= n; ++i) b[i] = a[i] - x;
  double res = 0, tmp = 0;
  for (int i = 1; i <= n; ++i) {
    if (tmp < 0) tmp = 0;
    tmp += b[i];
    res = max(res, tmp);
  }
  tmp = 0;
  for (int i = 1; i <= n; ++i) {
    if (tmp < 0) tmp = 0;
    tmp -= b[i];
    res = max(res, tmp);
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%lf", &a[i]);
  }
  while (r > eps + l) {
    double ll = (2 * l + r) / 3, rr = (l + 2 * r) / 3;
    double lll = solve(ll);
    double rrr = solve(rr);
    if (lll < rrr)
      r = rr;
    else
      l = ll;
  }
  cout << fixed << setprecision(12) << solve((l + r) / 2) << endl;
}