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 a1[200000];
int main() {
  int n;
  double l = -10000, r = 10000, md, a, b, c, d;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", a1 + i);
  while (r - l > 1e-12) {
    md = (l + r) / 2;
    a = -10001;
    b = 10001;
    c = d = 0;
    for (int i = 0; i < n; ++i) {
      c += a1[i] - md;
      if (c < 0) c = 0;
      if (c > a) a = c;
      d += a1[i] - md;
      if (d > 0) d = 0;
      if (d < b) b = d;
    }
    if (a + b > 0)
      l = md;
    else
      r = md;
  }
  printf("%.9f\n", a);
}