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 maxn = 2e5 + 5;
const int it_count = 150;
int a[maxn];
int n;
inline double f(double x) {
  double cur = 0, minn = 0, maxx = 0;
  double ret = 0.0;
  for (int pos = 0; pos < (int)n; pos++) {
    cur += a[pos] - x;
    minn = min(minn, cur);
    maxx = max(maxx, cur);
    ret = max(ret, max(abs(cur - minn), abs(cur - maxx)));
  }
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int j = 0; j < (int)n; j++) scanf("%d", &a[j]);
  double start = -2e4;
  double finish = 2e4;
  for (int iter = 0; iter < (int)it_count; iter++) {
    double m1 = start + (finish - start) / 3;
    double m2 = finish - (finish - start) / 3;
    double f1 = f(m1);
    double f2 = f(m2);
    if (f1 < f2)
      finish = m2;
    else
      start = m1;
  }
  double ans = f(start);
  printf("%.10lf\n", ans);
}