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 N;
double a[200200], g[200200], l, r, LT, RT, ans;
double f(double x) {
  double weakness = 0;
  for (int i = 0; i < (N); i++) g[i] = a[i] - x;
  double sum = 0;
  for (int i = 0; i < (N); i++) {
    sum = max(sum, 0.0);
    sum += g[i];
    weakness = max(weakness, sum);
  }
  sum = 0;
  for (int i = 0; i < (N); i++) {
    sum = min(sum, 0.0);
    sum += g[i];
    weakness = max(weakness, -sum);
  }
  return weakness;
}
int main() {
  scanf("%d", &N);
  for (int i = 0; i < (N); i++) {
    scanf("%lf", &a[i]);
  }
  l = -4e9, r = 4e9;
  while (true) {
    if (abs(r - l) < 1e-11) {
      ans = (r + l) / 2.0;
      break;
    }
    LT = l + (r - l) / 3.0;
    RT = r - (r - l) / 3.0;
    if (f(LT) > f(RT)) {
      l = LT;
    } else {
      r = RT;
    }
  }
  printf("%.12lf\n", f(ans));
  return 0;
}