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 s[204800], n;
double f(double x) {
  double l = 0, h = 0;
  for (int i = 1; i <= n; ++i) {
    l = min(l, i * x - s[i]);
    h = max(h, i * x - s[i]);
  }
  return h - l;
}
int main() {
  scanf("%d", &n);
  double L = 9e9, R = -9e9, t;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", s + i);
    s[i] += s[i - 1];
    t = s[i];
    t /= i;
    L = min(t, L);
    R = max(t, R);
  }
  for (int tm = 70; tm--;) {
    double M1 = (L * 2 + R) / 3, M2 = (L + R * 2) / 3;
    if (f(M1) < f(M2))
      R = M2;
    else
      L = M1;
  }
  printf("%.9lf\n", f(ldexp(L + R, -1)));
}