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 MAXN = 200200;
int n;
long long pre[MAXN];
long double arr[MAXN];
long double getval(long double x, long double s) {
  long double maxval = 0.0;
  long double cur = 0.0;
  for (int i = 0; i < n; i++) {
    long double v = arr[i] - x;
    v *= s;
    cur = max((long double)0.0, cur + v);
    maxval = max(cur, maxval);
  }
  return maxval;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int a;
    scanf("%d", &a);
    arr[i] = a;
    pre[i + 1] = pre[i] + a;
  }
  long double lo = -11000.0;
  long double hi = 11000.0;
  while (abs(hi - lo) > 1E-9) {
    long double mid = lo + (hi - lo) / 2;
    if (getval(mid, 1) > getval(mid, -1)) {
      lo = mid;
    } else {
      hi = mid;
    }
  }
  cout << setprecision(12) << fixed << getval(lo, -1) << endl;
  return 0;
}