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 INF = 987654321;
double calc(double x, vector<double> a) {
  int n = (int)a.size();
  for (int i = 0; i < n; ++i) a[i] -= x;
  double vmax = -1e18, vmin = 1e18;
  int p = 0, pp;
  while (p < n) {
    while (p < n && a[p] > 0.0) ++p;
    double cur = 0.0;
    pp = p;
    while (pp < n && cur + a[pp] <= 0.0) {
      cur += a[pp];
      vmin = min(vmin, cur);
      ++pp;
    }
    p = pp;
  }
  p = 0, pp;
  while (p < n) {
    while (p < n && a[p] < 0.0) ++p;
    double cur = 0.0;
    pp = p;
    while (pp < n && cur + a[pp] >= 0.0) {
      cur += a[pp];
      vmax = max(vmax, cur);
      ++pp;
    }
    p = pp;
  }
  return max(vmax, -vmin);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(12);
  cout << fixed;
  int n;
  cin >> n;
  vector<double> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  double l = -1e4, r = 1e4;
  for (int k = 0; k < 100; ++k) {
    double x1 = (2 * l + r) / 3;
    double x2 = (l + 2 * r) / 3;
    double v1 = calc(x1, a);
    double v2 = calc(x2, a);
    if (v1 < v2)
      r = x2;
    else
      l = x1;
  }
  cout << calc(l, a) << '\n';
  return 0;
}