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;
vector<long double> cum, cumt;
long double weakness() {
  long double mx = 0, mxc = 0, mn = INT_MAX, mnc = INT_MAX;
  mx = mxc = cum[0];
  for (int i = 1; i < n; i++) {
    mxc = max(cum[i], mxc + cum[i]);
    mx = max(mx, mxc);
  }
  mn = mnc = cum[0];
  for (int i = 1; i < n; i++) {
    mnc = min(cum[i], mnc + cum[i]);
    mn = min(mn, mnc);
  }
  return max(abs(mx), abs(mn));
}
long double calc(long double x) {
  for (double i = 0; i < n; i++) cum[i] = cumt[i] - x;
  return weakness();
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> n;
  cum.resize(n), cumt.resize(n);
  for (int i = 0; i < n; i++) cin >> cum[i];
  cumt = cum;
  long double a = -10000, d = 10000, b = (2 * a + d) / 3, c = (a + 2 * d) / 3;
  while (c - b > 0.000000000001) {
    if (calc(b) > calc(c))
      a = b;
    else
      d = c;
    b = (2 * a + d) / 3, c = (a + 2 * d) / 3;
  }
  cout << fixed << setprecision(7) << calc(b) << endl;
  cin.ignore(), cin.get();
}