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;
const int maxn = 2e5 + 5;
int a[maxn];
int p[maxn];
int n;
long double f(long double x) {
  long double minx = 0;
  long double maxx = 0;
  for (int i = 0; i <= n; i++)
    minx = min(minx, (long double)p[i] + (long double)i * x),
    maxx = max(maxx, (long double)p[i] + (long double)i * x);
  return maxx - minx;
}
int main() {
  scanf("%d", &n);
  p[0] = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    p[i + 1] = p[i] + a[i];
  }
  long double l = -3e9;
  long double r = 3e9;
  for (int it = 0; it < 150; it++) {
    long double m1 = l + (r - l) / 3.;
    long double m2 = r - (r - l) / 3.;
    if (f(m1) < f(m2))
      r = m2;
    else
      l = m1;
  }
  printf("%.30f", (double)(f(l)));
  return 0;
}