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 MX = 201020;
int N;
int ip[MX];
double ans = 1e9;
bool test(double x) {
  double minAns, minCur, maxAns, maxCur;
  minAns = maxAns = minCur = maxCur = 0.0;
  for (int i = 0; i < N; i++) {
    double t = ip[i] - x;
    minCur += t;
    maxCur += t;
    minCur = min(minCur, 0.0);
    maxCur = max(maxCur, 0.0);
    minAns = min(minAns, minCur);
    maxAns = max(maxAns, maxCur);
  }
  ans = min(ans, max(maxAns, -minAns));
  return maxAns >= (-minAns);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N;
  for (int i = 0; i < N; i++) {
    cin >> ip[i];
  }
  double l = -10000, r = 10000;
  while (l < r - 1e-11) {
    double md = (l + r) / 2.0;
    if (test(md)) {
      l = md;
    } else {
      r = md;
    }
  }
  test((l + r) / 2);
  cout << fixed << setprecision(10) << ans << endl;
  return 0;
}