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 N = 2e5 + 5;
double a[N], b[N], pr[N];
int n;
double f(double d) {
  for (int i = 1; i <= n; i++) {
    b[i] = a[i] - d;
    pr[i] = b[i] + pr[i - 1];
  }
  double ans = abs(b[n]);
  int mkmin = n, mkmax = n;
  for (int i = n - 1; i >= 1; i--) {
    if (pr[i] > pr[mkmax]) mkmax = i;
    if (pr[i] < pr[mkmin]) mkmin = i;
    ans = max(ans, abs(pr[mkmin] - pr[i - 1]));
    ans = max(ans, abs(pr[mkmax] - pr[i - 1]));
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  double l = -1e5, r = 1e5, mid1, mid2;
  for (int i = 1; i <= 300; i++) {
    mid1 = (2 * l + r) / 3;
    mid2 = (l + 2 * r) / 3;
    if (f(mid1) > f(mid2))
      l = mid1;
    else
      r = mid2;
  }
  l = (r + l) / 2;
  cout << fixed << setprecision(6) << f(l);
  return 0;
}