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 INF = 2000000000;
const int MOD = 1000000007;
long long Arr[200007], N;
double nz, ze;
double newa[200007];
const double EPS = 1E-12;
double process() {
  double nnew = 0, mxx = 0;
  for (int i = 0; i <= (N - 1); ++i) {
    nnew += newa[i];
    if (nnew > mxx)
      mxx = nnew;
    else if (nnew < 0)
      nnew = 0;
  }
  return mxx;
}
void calc(double val) {
  for (int i = 0; i <= (N - 1); ++i) newa[i] = double(Arr[i]) - val;
  nz = process();
  for (int i = 0; i <= (N - 1); ++i) newa[i] *= -1;
  ze = process();
}
int main(int argc, const char **argv) {
  ios::sync_with_stdio(false);
  while (cin >> N) {
    for (int i = 0; i <= (N - 1); ++i) cin >> Arr[i];
    double lo = *min_element(Arr, Arr + N);
    double hi = *max_element(Arr, Arr + N);
    int cnt = 0;
    while (cnt <= 200) {
      double mid = (hi + lo) / 2;
      calc(mid);
      if (nz > ze)
        lo = mid;
      else
        hi = mid;
      cnt++;
    }
    double mid = (hi + lo) / 2;
    calc(mid);
    cout << fixed << setprecision(10) << max(nz, ze) << endl;
  }
  return 0;
}