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;
double a[200005], ans;
int i, j, k, s, m, n;
inline double check(double o) {
  double ans1 = 0, ans2 = 0, ans = 0;
  for (int i = 1; i <= n; i++) {
    ans1 = ans1 + a[i] - o;
    ans2 = ans2 + a[i] - o;
    if (ans1 > 0) ans1 = 0;
    if (ans2 < 0) ans2 = 0;
    ans = max(ans, max(abs(ans1), abs(ans2)));
  }
  return ans;
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%lf", &a[i]);
  ans = n * 10000.0;
  double l = -10000.0, r = 10000.0;
  while (l + 1e-12 <= r) {
    double mid = (r - l) / 3.0 + l, mid1 = r - (r - l) / 3.0;
    double gt = check(mid), gtw = check(mid1);
    if (gt < gtw) {
      ans = min(ans, gt);
      r = mid1 - 1e-12;
    } else {
      ans = min(ans, gtw);
      l = mid + 1e-12;
    }
  }
  printf("%.9lf", ans);
}