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 maxi, mini, inf = 1e10, ma[200010], mi[200010], L, R, mid, eps = 1e-7,
                   now, las;
int n, a[200010], f = 0;
double work(double val) {
  maxi = -inf, mini = inf;
  for (int i = 1; i <= n; i++) {
    ma[i] = max(ma[i - 1], 0.0) + a[i] + val,
    mi[i] = min(mi[i - 1], 0.0) + a[i] + val;
    maxi = max(ma[i], maxi), mini = min(mi[i], mini);
  }
  if (abs(maxi) > abs(mini))
    return maxi;
  else
    return mini;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  L = -1e5, R = 1e5;
  while (1) {
    mid = (L + R) / 2;
    now = work(mid);
    if (f && abs(now - las) < eps) break;
    f = 1;
    las = now;
    if (now > 0)
      R = mid;
    else
      L = mid;
  }
  printf("%.6lf", abs(work(L)));
  return 0;
}