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;
void DBG() { cerr << "]" << endl; }
template <class H, class... T>
void DBG(H h, T... t) {
  cerr << h;
  if (sizeof...(t)) {
    cerr << ", ";
  }
  DBG(t...);
}
int n;
long double a[200005];
inline long double weakness(long double x) {
  long double s1 = 0.0, s2 = 0.0;
  long double mx = 0.0;
  for (int i = 0; i < n; i++) {
    long double temp = a[i] - x;
    s1 += temp;
    if (s1 < 0) s1 = 0;
    s2 += temp;
    if (s2 > 0) s2 = 0;
    mx = max(mx, max(s1, -s2));
  }
  return mx;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  long double l = -10001.0, r = 10001.0;
  while (r - l > 1e-12) {
    long double lm = (l * 2 + r) / 3;
    long double rm = (l + r * 2) / 3;
    if (weakness(lm) > weakness(rm))
      l = lm;
    else
      r = rm;
  }
  cout << setprecision(9) << weakness(l);
  return 0;
}