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 mn = 2e5 + 10;
int n;
long double a[mn], p[mn];
long double val(long double x) {
  long double sm = 0, lar = 0, ans = 0;
  for (int i = 1; i <= n; i++) {
    p[i] = p[i - 1] + a[i] - x;
    sm = min(sm, p[i]);
    lar = max(lar, p[i]);
    ans = max(ans, max(abs(p[i] - sm), abs(p[i] - lar)));
  }
  return ans;
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  long double l = -20000, r = 20000;
  while (r - l > 1e-12) {
    long double a = (2 * l + r) / 3, b = (2 * r + l) / 3;
    if (val(a) < val(b))
      r = b;
    else
      l = a;
  }
  printf("%.100Lf", val((l + r) / 2));
}