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 length;
vector<double> nums;
double kadane(double x) {
  double currentMax = nums[0] - x;
  double currentMin = nums[0] - x;
  double globalMax = nums[0] - x;
  double globalMin = nums[0] - x;
  for (unsigned i = 1; i < nums.size(); i++) {
    currentMax = max(nums[i] - x, nums[i] - x + currentMax);
    currentMin = min(nums[i] - x, nums[i] - x + currentMin);
    if (currentMax > globalMax) globalMax = currentMax;
    if (currentMin < globalMin) globalMin = currentMin;
  }
  if (abs(globalMin) < globalMax)
    return globalMax;
  else
    return abs(globalMin);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  double num;
  cin >> length;
  for (int i = 0; i < length; i++) {
    cin >> num;
    nums.push_back(num);
  }
  double l = -10000, r = 10000;
  double m1, m2;
  double resM1, resM2;
  for (int i = 0; i < 200; i++) {
    m1 = l + (r - l) / 3;
    m2 = r - (r - l) / 3;
    resM1 = kadane(m1);
    resM2 = kadane(m2);
    if (resM1 < resM2)
      r = m2;
    else
      l = m1;
  }
  cout << fixed << setprecision(7) << resM1 << endl;
  return 0;
}