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 n, a[(int)2e5 + 10];
long double WeekNess, mx = -1e18, mn = 1e18, sum;
long double Answer(long double Check) {
  mx = 0, mn = 0, sum = 0, WeekNess = 0;
  for (int i = 0; i < n; i++) {
    mx = max(mx, sum);
    mn = min(mn, sum);
    sum += a[i] - Check;
    WeekNess = max(WeekNess, fabs(sum - mx));
    WeekNess = max(WeekNess, fabs(sum - mn));
  }
  return WeekNess;
}
long double BinarySol(long double l, long double r) {
  long double mid1 = (2 * l + r) / 3;
  long double mid2 = (2 * r + l) / 3;
  int c = 0;
  while (c++ <= 200) {
    mid1 = (2 * l + r) / 3;
    mid2 = (2 * r + l) / 3;
    if (Answer(mid1) >= Answer(mid2)) {
      l = mid1;
    } else {
      r = mid2;
    }
  }
  return l;
}
int main() {
  cin >> n;
  for (int i = 0; i < n && scanf("%d", a + i); i++)
    ;
  cout << setprecision(6) << fixed << Answer(BinarySol(-1e10, 1e10));
}