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;
int a[200002];
pair<double, int> weakness(double x) {
  double mini = 0, maxi = 0, p = 0, g = 0;
  for (int i = 1; i < n + 1; i++) {
    if (mini > a[i] - x * i) {
      p = i;
      mini = a[i] - x * i;
    }
    if (maxi < a[i] - x * i) {
      g = i;
      maxi = a[i] - x * i;
    }
  }
  return make_pair(maxi - mini, g - p);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n + 1; i++) scanf("%d", &a[i]);
  for (int i = 2; i < n + 2; i++) a[i] += a[i - 1];
  double l = -1000000000, r = 1000000000;
  pair<double, int> wt;
  for (int i = 0; i < 80; i++) {
    wt = weakness((l + r) / 2);
    if (wt.second == 0) {
      printf("0\n");
      return 0;
    }
    if (wt.second > 0)
      l = (l + r) / 2;
    else
      r = (l + r) / 2;
  }
  printf("%.7lf\n", wt.first);
}