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 IN() {
  int c, f, x;
  while (!isdigit(c = getchar()) && c != '-')
    ;
  c == '-' ? (f = 1, x = 0) : (f = 0, x = c - '0');
  while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + c - '0';
  return !f ? x : -x;
}
const int N = 200000 + 19;
int A[N], n;
double B[N], l, r, res, tmp;
double calc(double x) {
  res = 0;
  for (int i = 1; i < n + 1; i++) B[i] = A[i] - x;
  tmp = 0;
  for (int i = 1; i < n + 1; i++) {
    tmp = max(tmp, 0.0) + B[i];
    res = max(res, abs(tmp));
  }
  tmp = 0;
  for (int i = 1; i < n + 1; i++) {
    tmp = min(tmp, 0.0) + B[i];
    res = max(res, abs(tmp));
  }
  return res;
}
int main() {
  n = IN();
  for (int i = 1; i < n + 1; i++) A[i] = IN();
  l = -10000, r = 10000;
  for (int i = 0; i < 100; i++) {
    double t = (r - l) / 3.0;
    double t1 = l + t, t2 = r - t;
    if (calc(t1) > calc(t2))
      l = t1;
    else
      r = t2;
  }
  printf("%.10lf\n", calc(l));
}