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 p, q, r, n, a[200005];
double eps = 5e-7;
double mid(double a, double b) {
  if (a > b) swap(a, b);
  return a + (b - a) * 0.618;
}
double fun(double x) {
  double MAX, MIN, sum;
  int i;
  MAX = MIN = sum = 0;
  for (i = 0; i < n; i++) {
    sum += a[i] - x;
    MAX = max(MAX, sum);
    MIN = min(MIN, sum);
  }
  return MAX - MIN;
}
double dis(double a, double b) { return fabs(a - b); }
double find() {
  double m1, m2, l, r;
  double cl, cr, cm1, cm2;
  r = 100000;
  l = -100000;
  cl = fun(l);
  cr = fun(r);
  m2 = mid(r, l);
  cm2 = fun(m2);
  m1 = mid(l, m2);
  cm1 = fun(m1);
  while (fabs(l - r) + fabs(cl - cr) > eps) {
    if (cm1 < cm2) {
      r = m2;
      cr = cm2;
      m2 = m1;
      cm2 = cm1;
      m1 = mid(l, m2);
      cm1 = fun(m1);
    } else {
      l = m1;
      cl = cm1;
      m1 = m2;
      cm1 = cm2;
      m2 = mid(r, l);
      cm2 = fun(m2);
    }
  }
  return (cm1 + cm2) / 2;
}
int main() {
  int t, i;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d", &a[i]);
  printf("%.10f\n", find());
  return 0;
}