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 a[300005];
int n;
long double ans = 10000000000;
long double test(long double x) {
  long double sum = 0;
  long double mn = 0;
  long double mx = 0;
  long double ret = 0;
  for (int i = 0; i < n; i++) {
    sum += (a[i] - x);
    ret = max(ret, abs(sum - mn));
    ret = max(ret, abs(sum - mx));
    mn = min(mn, sum);
    mx = max(mx, sum);
  }
  ans = min(ans, ret);
  return ret;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  long double s = -10000000, e = 10000000;
  while (e - s > 1e-12) {
    long double t1 = s + (e - s) / 3;
    long double t2 = s + (e - s) * 2 / 3;
    long double v1 = test(t1);
    long double v2 = test(t2);
    if (v1 < v2) {
      e = t2;
    } else {
      s = t1;
    }
  }
  printf("%.9Lf", ans);
}