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 SIZE = 2e5;
int a[SIZE], n;
double f(double x) {
  double ma = 0, mi = 0, now = 0;
  double res = 0;
  for (int i = 1; i <= n; i++) {
    now += a[i] - x;
    res = max(res, fabs(ma - now));
    res = max(res, fabs(mi - now));
    ma = max(ma, now);
    mi = min(mi, now);
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  double ll = -10000, rr = 10000;
  for (int k = 0; k < 100; k++) {
    double mm1 = ll * 0.54 + rr * 0.46;
    double mm2 = ll * 0.46 + rr * 0.54;
    double v1 = f(mm1);
    double v2 = f(mm2);
    if (v1 > v2)
      ll = mm1;
    else
      rr = mm2;
  }
  printf("%.15f\n", f((ll + rr) * 0.5));
  return 0;
}