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;
inline int in(int d = 0, char q = 0, int c = 1) {
  while (q != '-' && q != EOF && (q < 48 || q > 57)) q = getchar();
  if (q == EOF) return EOF;
  if (q == '-') c = -1, q = getchar();
  do d = d * 10 + (q ^ 48), q = getchar();
  while (q < 58 && q > 47);
  return c * d;
}
int main() {
  int n = in(), a[200000], cnt = 0;
  double Ma = -10000000, Mi = 10000000, EPS = 1E-9, R = -10000, L = 10000, M,
         q1, q2;
  for (int i = 0; i < n; i++) {
    a[i] = in();
  }
  while (fabs(Mi - Ma) > EPS && cnt < 100) {
    cnt++;
    M = (R + L) / 2.0;
    q1 = q2 = 0;
    Ma = Mi = 0;
    for (int i = 0; i < n; i++) {
      q1 += -M + a[i];
      q2 += -M + a[i];
      if (q1 < 0) q1 = 0;
      if (q2 > 0) q2 = 0;
      Ma = max(Ma, q1);
      Mi = min(Mi, q2);
    }
    if (Ma > -Mi)
      R = M;
    else
      L = M;
  }
  printf("%.7lf\n", max(fabs(Mi), fabs(Ma)));
}