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 n, a[200001];
long double b[200001], pre[200001], ans = 2e11;
long double check(long double x) {
  for (int i = 1; i <= n; i++) b[i] = a[i] - x;
  for (int i = 1; i <= n; i++) pre[i] = pre[i - 1] + b[i];
  long double mn = 2e13, mx = -2e13;
  long double mns = 0, mxs = 0;
  for (int i = 1; i <= n; i++) {
    mn = min(mn, pre[i] - mxs);
    mx = max(mx, pre[i] - mns);
    mns = min(mns, pre[i]);
    mxs = max(mxs, pre[i]);
  }
  return max(-mn, mx);
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  long double low = -1e4, high = 1e4;
  for (int i = 0; i <= 120; i++) {
    long double med = (low + high) / 2;
    long double x = check(med);
    long double y = check(med + 1e-11);
    ans = min(ans, min(x, y));
    if (x < y)
      high = med;
    else
      low = med + 1e-11;
  }
  cout << fixed << setprecision(8) << ans << '\n';
}