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 M = 200005;
int A[M];
double B[M];
int n;
double check(double x) {
  for (int i = 1, i_END_ = n + 1; i < i_END_; ++i) B[i] = A[i] - x;
  double mx = -1e60, mi = 1e60, s1 = 0, s2 = 0;
  for (int i = 1, i_END_ = n + 1; i < i_END_; ++i) {
    s1 += B[i];
    mx = max(s1, mx);
    if (s1 < 0) s1 = 0;
    s2 += B[i];
    mi = min(mi, s2);
    if (s2 > 0) s2 = 0;
  }
  return max(mx, -mi);
}
int main() {
  scanf("%d", &n);
  for (int i = 1, i_END_ = n + 1; i < i_END_; ++i) scanf("%d", &A[i]);
  double L = -10000, R = 10000, res = -10001;
  double ans = 1e60;
  for (int i = 0, i_END_ = 100; i < i_END_; ++i) {
    double del = (R - L) / 3;
    double m1 = L + del, m2 = R - del;
    double r1 = check(m1), r2 = check(m2);
    ans = min(ans, min(r1, r2));
    if (r2 < r1)
      L = m1;
    else
      R = m2;
  }
  printf("%.12f\n", ans);
  return 0;
}