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 A[200050];
double L = -1e6, R = 1e6;
double ans = (1e9);
int N;
double f(double x) {
  double maxm = 0;
  double minm = 0;
  double pre = 0;
  double ans = -1000000009;
  for (int i = 1; i <= N; i++) {
    pre += A[i] - x;
    ans = max(ans, abs(pre - maxm));
    ans = max(ans, abs(pre - minm));
    maxm = max(maxm, pre);
    minm = min(minm, pre);
  }
  return ans;
}
int main() {
  cin.tie(0), ios::sync_with_stdio(false);
  cin >> N;
  for (int i = 1; i <= N; i++) cin >> A[i];
  for (int i = 1; i <= 100; i++) {
    double mid1 = L + (R - L) / (3.0);
    double mid2 = R - (R - L) / (3.0);
    double x = f(mid1), y = f(mid2);
    ans = min(ans, min(x, y));
    if (x < y)
      R = mid2;
    else
      L = mid1;
  }
  cout << fixed << setprecision(15) << ans << endl;
}