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 double EXP = 1e-11;
const int maxn = 2e5 + 10;
int n;
double a[maxn];
double a2[maxn];
double Judge(double x) {
  double sum1 = 0.0, sum2 = 0.0, sum3 = 0.0;
  for (int i = 1; i <= n; i++) a2[i] = a[i] - x;
  for (int i = 1; i <= n; i++) {
    sum1 = max(sum1 + a2[i], a2[i]);
    sum3 = max(sum3, sum1);
  }
  for (int i = 1; i <= n; i++) a2[i] = a2[i] * (-1.0);
  for (int i = 1; i <= n; i++) {
    sum2 = max(sum2 + a2[i], a2[i]);
    sum3 = max(sum3, sum2);
  }
  return sum3;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lf", &a[i]);
  double l = -40000.0, r = 40000.0;
  while (r - l > EXP) {
    double mid = (l + l + r) / 3.0;
    double mmid = (l + r + r) / 3.0;
    if (Judge(mid) >= Judge(mmid))
      l = mid;
    else
      r = mmid;
  }
  printf("%.13lf\n", min(Judge(l), Judge(r)));
  return 0;
}