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[200005];
double contisum() {
  double maxx = 0, cur = 0;
  for (int i = 0; i < n; i++) {
    if (cur > -1e-8)
      cur += a[i];
    else
      cur = a[i];
    if (cur - maxx > 1e-8) maxx = cur;
  }
  return maxx;
}
double check(double x) {
  for (int i = 0; i < n; i++) a[i] -= x;
  double maxx1 = contisum();
  for (int i = 0; i < n; i++) a[i] = -a[i];
  double maxx2 = contisum();
  for (int i = 0; i < n; i++) a[i] = -a[i];
  for (int i = 0; i < n; i++) a[i] += x;
  if (maxx1 - maxx2 > 1e-8)
    return maxx1;
  else
    return maxx2;
}
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 0; i < n; i++) scanf("%lf", &a[i]);
    double ans = 0x7fffffff;
    double l = -10000.0, r = 10000.0;
    for (int i = 0; i < 200; i++) {
      double mid = (l + r) / 2;
      double rmid = (mid + r) / 2;
      double temp1 = check(mid);
      double temp2 = check(rmid);
      if (temp1 - temp2 < 1e-8)
        r = rmid;
      else
        l = mid;
      ans = min(ans, min(temp1, temp2));
    }
    printf("%.12f\n", ans);
  }
  return 0;
}