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[200000];
int main() {
  int N;
  scanf("%d", &N);
  for (int i = 0; i < N; i++) scanf("%d", &A[i]);
  long double mx, mn, l = -10000, r = 10000, m;
  while (r - l > 1e-13) {
    m = (l + r) / 2;
    mx = -1e20, mn = 1e20;
    long double tx = 0, tn = 0;
    for (int i = 0; i < N; i++) {
      mx = max(mx, tx = A[i] - m + max(0.l, tx));
      mn = min(mn, tn = A[i] - m + min(0.l, tn));
    }
    mx + mn > 0 ? l = m : r = m;
  }
  printf("%.8f", (double)mx);
  return 0;
}