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;
vector<long long> A;
void calc(double x, double& pos, double& neg) {
  double max_ending_here = (A[0] - x);
  double max_from_start = (A[0] - x);
  for (int i = 1; i < A.size(); i++) {
    if (max_ending_here + (A[i] - x) < (A[i] - x)) {
      max_ending_here = (A[i] - x);
    } else {
      max_ending_here += (A[i] - x);
    }
    max_from_start = max(max_from_start, max_ending_here);
  }
  pos = max_from_start;
  max_ending_here = -(A[0] - x);
  max_from_start = -(A[0] - x);
  for (int i = 1; i < A.size(); i++) {
    if (max_ending_here + (-(A[i] - x)) < (-(A[i] - x))) {
      max_ending_here = (-(A[i] - x));
    } else {
      max_ending_here += (-(A[i] - x));
    }
    max_from_start = max(max_from_start, max_ending_here);
  }
  neg = max_from_start;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int t;
    scanf("%d", &t);
    A.push_back(t);
  }
  double lo = -11111, hi = 11111;
  for (int i = 0; i < 100; i++) {
    double a = (lo * 2 + hi) / 3;
    double b = (lo + hi * 2) / 3;
    double t, s;
    calc(a, t, s);
    double aa = max(t, s);
    calc(b, t, s);
    double bb = max(t, s);
    if (aa < bb) {
      hi = b;
    } else {
      lo = a;
    }
  }
  double ans1, ans2;
  calc((lo + hi) / 2, ans1, ans2);
  printf("%.8lf\n", max(ans1, ans2));
}