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 N = 200200;
double a[N];
int n, m, i, j, k;
double l, r, f1, f2, u, v;
double solve(double c) {
  int i;
  double sum, ans, l, r;
  l = r = sum = 0;
  for (i = 1; i <= n; i++) {
    sum += a[i] - c;
    l = min(l, sum);
    r = max(r, sum);
  }
  return r - l;
}
int main() {
  scanf("%d", &n);
  l = 1e8;
  r = -1e8;
  for (i = 1; i <= n; i++) {
    scanf("%lf", &a[i]);
    r = max(r, a[i]);
    l = min(l, a[i]);
  }
  while (1) {
    u = l + (r - l) / 3;
    v = l + (r - l) / 3 * 2;
    f1 = solve(u);
    f2 = solve(v);
    if (fabs(f1 - f2) <= 1e-8 && fabs(r - l) <= 1e-11) break;
    if (f1 <= f2)
      r = v;
    else
      l = u;
  }
  printf("%.9lf\n", f1);
  return 0;
}