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;
long long int modpow(long long int a, long long int n, long long int temp) {
  long long int res = 1, y = a;
  while (n > 0) {
    if (n & 1) res = (res * y) % temp;
    y = (y * y) % temp;
    n /= 2;
  }
  return res % temp;
}
vector<double> c;
double mcs(vector<int> &a, double x) {
  int i;
  for (i = 0; i < c.size(); ++i) c[i] = a[i] - x;
  double sum = 0, res = 0;
  for (i = 0; i < c.size(); ++i) {
    sum += c[i];
    res = max(res, sum);
    if (sum < 0) sum = 0;
  }
  return res;
}
int main() {
  int n, i;
  cin >> n;
  c = vector<double>(n);
  vector<int> a(n), b(n);
  for (i = 0; i < n; ++i) {
    cin >> a[i];
    b[i] = -a[i];
  }
  double l = *min_element(a.begin(), a.end()),
         r = *max_element(a.begin(), a.end());
  while (fabs(l - r) > 1e-11) {
    double m = (l + r) / 2;
    double pos = mcs(a, m);
    double neg = mcs(b, -m);
    if (pos > neg) {
      l = m;
    } else {
      r = m;
    }
  }
  double m = (l + r) / 2;
  printf("%.7lf", max(mcs(a, m), mcs(b, -m)));
  return 0;
}