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;
long double v[200101];
const long double eps = 1e-8;
int n;
long double f(long double x) {
  long double maxx = -1e18;
  long double minx = 1e18;
  long double last_max = 0;
  long double last_min = 0;
  for (int i = 0; i < n; i++) {
    last_max += v[i] - x;
    last_min += v[i] - x;
    maxx = max(maxx, last_max);
    minx = min(minx, last_min);
    if (last_max + eps < 0.0) {
      last_max = 0;
    }
    if (last_min - eps > 0.0) {
      last_min = 0;
    }
  }
  return max(fabs(maxx), fabs(minx));
}
int main() {
  scanf("%d", &n);
  double sum = 0;
  double b1 = -1e18;
  double b2 = 1e18;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    sum += v[i];
    b1 = max(b1, sum / (i + 1));
    b2 = min(b2, sum / (i + 1));
  }
  long double l = -5e4, r = 5e4;
  while (l + eps < r) {
    long double m1 = l + (r - l) / 3.0;
    long double m2 = r - (r - l) / 3.0;
    long double f1 = f(m1);
    long double f2 = f(m2);
    if (f1 - eps > f2) {
      l = m1;
    } else {
      r = m2;
    }
  }
  printf("%.7lf", (double)min(f(l), min(f(b1), f(b2))));
  return 0;
}