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 a[200100], s[200100], v1[200100], v2[200100], sz1, sz2;
int main() {
  double mx = 1e20, ans;
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    s[i] = s[i - 1] + a[i];
  }
  for (int i = 0; i <= n; i++) {
    while (sz1 >= 2 &&
           1LL * (s[v1[sz1 - 1]] - s[v1[sz1 - 2]]) * (i - v1[sz1 - 2]) >=
               1LL * (s[i] - s[v1[sz1 - 2]]) * (v1[sz1 - 1] - v1[sz1 - 2]))
      sz1--;
    v1[sz1++] = i;
  }
  for (int i = n; i >= 0; i--) {
    while (sz2 >= 2 &&
           1LL * (s[v2[sz2 - 1]] - s[v2[sz2 - 2]]) * (i - v2[sz2 - 2]) >=
               1LL * (s[i] - s[v2[sz2 - 2]]) * (v2[sz2 - 1] - v2[sz2 - 2]))
      sz2--;
    v2[sz2++] = i;
  }
  for (int i = 0, j = 0; i < sz1 - 1 || j < sz2 - 1;) {
    double nx1 =
        i < sz1 - 1 ? 1.0 * (s[v1[i + 1]] - s[v1[i]]) / (v1[i + 1] - v1[i]) : 0;
    double nx2 =
        j < sz2 - 1 ? 1.0 * (s[v2[j + 1]] - s[v2[j]]) / (v2[j + 1] - v2[j]) : 0;
    double nx;
    if (i < sz1 - 1 && j < sz2 - 1 && nx1 < nx2 || j == sz2 - 1) {
      i++;
      nx = nx1;
    } else {
      j++;
      nx = nx2;
    }
    double ny1 = s[v1[i]] - nx * v1[i];
    double ny2 = s[v2[j]] - nx * v2[j];
    if (ny2 - ny1 < mx) {
      mx = ny2 - ny1;
      ans = nx;
    }
  }
  printf("%.8f\n", mx);
  return 0;
}