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;
vector<int> a;
vector<int> acomp;
double get(double mid) {
  double ans1 = INT_MIN;
  double ans2 = INT_MIN;
  double sum1 = 0;
  double sum2 = 0;
  for (int i = 0; i < a.size(); i++) {
    sum1 = sum1 + a[i] - mid;
    sum2 = sum2 + acomp[i] + mid;
    if (sum1 > ans1) {
      ans1 = sum1;
    }
    if (sum2 > ans2) {
      ans2 = sum2;
    }
    if (sum1 < 0) {
      sum1 = 0;
    }
    if (sum2 < 0) {
      sum2 = 0;
    }
  }
  return max(ans1, ans2);
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    a.push_back(x);
    acomp.push_back(-x);
  }
  double l = -1e9;
  double r = 1e9;
  while (r - l > 1e-11) {
    double mid1 = (r + 2 * l) / 3;
    double mid2 = (l + 2 * r) / 3;
    double val1 = get(mid1);
    double val2 = get(mid2);
    if (val1 > val2) {
      l = mid1;
    } else {
      r = mid2;
    }
  }
  printf("%0.12lg\n", min(get(l), get(r)));
}