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;
vector<int> v(2 * 100000);
vector<double> vv(2 * 100000);
inline void calc(int l, double m, double &p, double &n) {
  int x;
  for (x = 0; x < l; x++) vv[x] = v[x] * 1.0 - m;
  double mx = 0, sm = 0;
  for (x = 0; x < l; x++) {
    sm += vv[x];
    if (sm > mx) mx = sm;
    if (sm < 0) sm = 0;
  }
  p = mx;
  for (x = 0; x < l; x++) vv[x] = -vv[x];
  mx = sm = 0;
  for (x = 0; x < l; x++) {
    sm += vv[x];
    if (sm > mx) mx = sm;
    if (sm < 0) sm = 0;
  }
  n = mx;
}
int main() {
  scanf("%d", &n);
  ;
  int x;
  for (x = 0; x < n; x++) scanf("%d", &v[x]);
  ;
  double l = *min_element(v.begin(), v.begin() + n);
  double r = *max_element(v.begin(), v.begin() + n);
  double eps = 1e-6;
  double ans;
  for (x = 1; x <= 100; x++) {
    double m = (l + r) / 2.0;
    double pos, neg;
    calc(n, m, pos, neg);
    if (pos > neg)
      l = m;
    else if (pos < neg)
      r = m;
    else if (abs(pos - neg) == eps) {
      ans = pos;
      break;
    }
  }
  printf("%.7lf", ans);
}