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 N = 2e5 + 5;
int n;
double a[N];
double max_poor(double x) {
  int i;
  double pre, mpre, ret = 0;
  pre = mpre = 0;
  for (i = 0; i < n; i++) {
    pre += a[i] - x;
    ret = max(ret, abs(pre - mpre));
    mpre = min(mpre, pre);
  }
  pre = mpre = 0;
  for (i = 0; i < n; i++) {
    pre += a[i] - x;
    ret = max(ret, abs(pre - mpre));
    mpre = max(mpre, pre);
  }
  return ret;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(0);
  int i;
  double l = 0.0, h = 0.0, m1, m2;
  cin >> n;
  for (i = 0; i < n; h = max(h, abs(a[i])), i++) cin >> a[i];
  l = -h;
  for (i = 0; i < 300; i++) {
    m1 = l + (h - l) / 3;
    m2 = h - (h - l) / 3;
    if (max_poor(m1) > max_poor(m2))
      l = m1;
    else
      h = m2;
  }
  printf("%.17f\n", max_poor(l));
  return 0;
}