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 MAXN = 200000;
using namespace std;
void RI() {}
template <typename... T>
void RI(int& head, T&... tail) {
  scanf("%d", &head);
  RI(tail...);
}
int a[200000];
vector<pair<double, double> > lines;
double intersection(pair<double, double> a, pair<double, double> b) {
  double x = (a.second - b.second) / (a.first - b.first);
  double y = (a.first * x) - a.second;
  return x;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  RI(n);
  for (int i = 0; i < int(n); i++) RI(a[i]);
  double c = 0;
  lines.push_back(make_pair(0, 0));
  for (int i = 0; i < int(n); i++) {
    c = c + a[i];
    lines.push_back(make_pair(i + 1, c));
  }
  vector<int> s1;
  vector<int> s2;
  s1.push_back(n);
  s1.push_back(n - 1);
  for (int j = n - 2; j >= 0; j--) {
    while (1) {
      if (s1.size() == 1) {
        break;
      }
      if (intersection(lines[s1[s1.size() - 2]], lines[j]) >=
          intersection(lines[s1[s1.size() - 2]], lines[s1[s1.size() - 1]])) {
        s1.pop_back();
      } else {
        break;
      }
    }
    s1.push_back(j);
  }
  s2.push_back(0);
  s2.push_back(1);
  for (int j = 2; j <= n; j++) {
    while (1) {
      if (s2.size() == 1) {
        break;
      }
      if (intersection(lines[s2[s2.size() - 2]], lines[j]) >=
          intersection(lines[s2[s2.size() - 2]], lines[s2[s2.size() - 1]])) {
        s2.pop_back();
      } else {
        break;
      }
    }
    s2.push_back(j);
  }
  int curr_line = 0;
  double ans = 1000000000.0;
  for (int j = 0; j < s2.size() - 1; j++) {
    double inter1 = intersection(lines[s2[j]], lines[s2[j + 1]]);
    while (1) {
      if (curr_line >= s1.size() - 1) {
        break;
      }
      double inter =
          intersection(lines[s1[curr_line]], lines[s1[curr_line + 1]]);
      if (inter <= inter1) {
        break;
      } else {
        curr_line++;
      }
    }
    double y1 = (lines[s2[j]].first * inter1) - lines[s2[j]].second;
    double y2 =
        (lines[s1[curr_line]].first * inter1) - lines[s1[curr_line]].second;
    ans = min(abs(y2 - y1), ans);
  }
  curr_line = 0;
  for (int j = 0; j < s1.size() - 1; j++) {
    double inter1 = intersection(lines[s1[j]], lines[s1[j + 1]]);
    while (1) {
      if (curr_line >= s2.size() - 1) {
        break;
      }
      double inter =
          intersection(lines[s2[curr_line]], lines[s2[curr_line + 1]]);
      if (inter <= inter1) {
        break;
      } else {
        curr_line++;
      }
    }
    double y1 = (lines[s1[j]].first * inter1) - lines[s1[j]].second;
    double y2 =
        (lines[s2[curr_line]].first * inter1) - lines[s2[curr_line]].second;
    ans = min(abs(y2 - y1), ans);
  }
  printf("%.12f\n", ans);
  return 0;
}