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>
const double eps = 1e-10;
const int MaxN = 200005;
double a[MaxN];
int n;
double sol(double x) {
  double s = 0, mx = 0, mn = 0, res = -1e20;
  for (int i = 0; i < n; i++) {
    s += a[i] - x;
    res = std::max(res, fabs(s - mn));
    res = std::max(res, fabs(s - mx));
    mx = std::max(mx, s);
    mn = std::min(mn, s);
  }
  return res;
}
int main() {
  while (scanf("%d", &n) == 1) {
    for (int i = 0; i < n; i++) {
      scanf("%lf", &a[i]);
    }
    double l = -1e20, r = 1e20, mid1, mid2, res1, res2;
    for (int i = 0; i < 200; i++) {
      mid1 = (2 * l + r) / 3.0;
      mid2 = (l + 2 * r) / 3.0;
      res1 = sol(mid1);
      res2 = sol(mid2);
      if (res1 < res2 + eps) {
        r = mid2;
      } else {
        l = mid1;
      }
    }
    printf("%.12f\n", sol((l + r) / 2.0));
  }
  return 0;
}