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 = 1e6 + 7;
const long double EPS = 1e-12;
int mn, mx, n;
long double p[N];
long double f(long double x) {
  mn = mx = 0;
  for (int i = 1; i <= n; i++) {
    if (p[i] - x * i > p[mx] - x * mx) mx = i;
    if (p[i] - x * i < p[mn] - x * mn) mn = i;
  }
  return (p[mx] - x * mx) - (p[mn] - x * mn);
}
long double l, r, m1, m2, f1, f2;
int a[N];
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    p[i + 1] = p[i] + a[i];
  }
  l = -120000;
  r = 120000;
  while (clock() < 1200) {
    m1 = (l + l + r) / 3.0;
    m2 = (l + r + r) / 3.0;
    f1 = f(m1);
    f2 = f(m2);
    if (f1 < f2)
      r = m2;
    else
      l = m1;
  }
  cout.precision(10);
  cout << fixed << f((l + r) / 2.0);
  return 0;
}