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;
void FastIO() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
}
int n;
vector<int> a;
double foo(double val) {
  double ret1 = -INT_MAX, ret2 = -INT_MAX, temp1 = 0, temp2 = 0;
  for (int i = 0; i < (int)n; i++) {
    temp1 += a[i] - val;
    temp1 *= (temp1 >= 0);
    ret1 = max(ret1, temp1);
    temp2 += a[i] - val;
    temp2 *= (temp2 <= 0);
    ret2 = max(ret2, fabs(temp2));
  }
  return max(ret1, ret2);
}
int main() {
  FastIO();
  cin >> n;
  a.assign(n, 0);
  for (int i = 0; i < (int)n; i++) {
    cin >> a[i];
  }
  double l = -20000, r = 20000, mid1, mid2, cnt = 400, ans = 0;
  while (cnt--) {
    mid1 = l + (r - l) / 3;
    mid2 = r - ((r - l) / 3);
    double val1 = foo(mid1);
    double val2 = foo(mid2);
    if (mid1 == mid2) {
      ans = val1;
      break;
    } else if (val1 < val2) {
      r = mid2;
    } else {
      l = mid1;
    }
  }
  cout << setprecision(10) << fixed << ans << '\n';
  return 0;
}