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 main() {
  int n;
  cin >> n;
  vector<int> a(n), b(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  double l = *min_element(a.begin(), a.end());
  double r = *max_element(a.begin(), a.end());
  double mx = 0, mn = 0;
  while (r - l > 1e-12) {
    double mid = (l + r) / 2, sum1 = 0, sum2 = 0;
    mx = mn = 0;
    for (int i = 0; i < n; i++) {
      sum1 += a[i] - mid;
      if (sum1 < 0) sum1 = 0;
      if (sum1 > mx) mx = sum1;
      sum2 += a[i] - mid;
      if (sum2 > 0) sum2 = 0;
      if (sum2 < mn) mn = sum2;
    }
    if (mx + mn > 0)
      l = mid;
    else
      r = mid;
  }
  printf("%.12f\n", max(mx, abs(mn)));
  return 0;
}