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[200005];
double fmin(double x, int n) {
  double ans = 0, cur = 0;
  for (int i = 0; i < n; i++) {
    cur += (a[i] - x);
    if (cur > 0) cur = 0;
    ans = min(ans, cur);
  }
  return ((ans) > 0 ? (ans) : (-(ans)));
}
double fmax(double x, int n) {
  double ans = 0, cur = 0;
  for (int i = 0; i < n; i++) {
    cur += (a[i] - x);
    if (cur < 0) cur = 0;
    ans = max(ans, cur);
  }
  return ((ans) > 0 ? (ans) : (-(ans)));
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  double lo = -10000, hi = 10000, mid, v1 = -200, v2 = 200;
  while (((v1 - v2) > 0 ? (v1 - v2) : (-(v1 - v2))) > .4 * 1e-6) {
    mid = (lo + hi) / 2;
    v1 = fmin(mid, n);
    v2 = fmax(mid, n);
    if (v1 < v2)
      lo = mid;
    else
      hi = mid;
  }
  printf("%.9lf\n", v1);
  return 0;
}