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;
double ans = 0;
double x = 0;
int n;
vector<double> v;
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  v.resize(n + 1);
  for (int i = 1; i <= n; i++) cin >> v[i];
  double ans = 0.00000000;
  double L = -100005, R = -L;
  for (int roda = 0; roda <= 90; roda++) {
    double mid = (L + R) / 2.000;
    double mxans = 0.0000000000000, mnans = 0.00000000000;
    double lxans = 0.000000000000, lyans = 0.00000000000;
    double sj = 0.00;
    for (int i = 1; i <= n; i++) {
      sj += v[i] - mid;
      double lans = sj, rans = sj;
      lans = sj - mxans;
      rans = sj - mnans;
      if (i == 1) {
        lxans = sj;
        lyans = sj;
        mxans = sj;
        mnans = sj;
      } else {
        lxans = min(lxans, lans);
        lyans = max(lyans, rans);
        lxans = min(lxans, sj);
        lyans = max(lyans, sj);
        mxans = max(mxans, sj);
        mnans = min(mnans, sj);
      }
    }
    if (-lxans > lyans) {
      R = mid;
    } else
      L = mid;
  }
  double mid = (L + R) / 2.000;
  double mxans = 0.0000000000000, mnans = 0.00000000000;
  double lxans = 0.000000000000, lyans = 0.00000000000;
  double sj = 0.00;
  for (int i = 1; i <= n; i++) {
    sj += v[i] - mid;
    double lans = sj, rans = sj;
    lans = sj - mxans;
    rans = sj - mnans;
    if (i == 1) {
      lxans = sj;
      lyans = sj;
      mxans = sj;
      mnans = sj;
    } else {
      lxans = min(lxans, lans);
      lyans = max(lyans, rans);
      lxans = min(lxans, sj);
      lyans = max(lyans, sj);
      mxans = max(mxans, sj);
      mnans = min(mnans, sj);
    }
  }
  ans = max(fabs(lyans), fabs(lxans));
  cout << fixed << setprecision(9) << ans << endl;
}