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;
long long MOD = 1000000007;
long double solve(vector<long long>& v, long double x) {
  long double mn = 0;
  long double mx = 0;
  long double ans = 0;
  for (int i = 0; i < (int)v.size(); ++i) {
    long double cur = 1.0 * v[i] - x * (i + 1);
    ans = max(ans, abs(cur - mx));
    ans = max(ans, abs(cur - mn));
    mx = max(mx, cur);
    mn = min(mn, cur);
  }
  return ans;
}
int main() {
  int n;
  cin >> n;
  vector<long long> v(n);
  for (int i = 0; i < (n); ++i) {
    scanf("%lld", &v[i]);
    if (i) v[i] += v[i - 1];
  }
  long double l = -3e4, r = +3e4, mid1, mid2;
  int iter = 100;
  while (iter--) {
    mid1 = (r - l) / 3.0;
    mid2 = r - mid1;
    mid1 += l;
    long double v1 = solve(v, mid1);
    long double v2 = solve(v, mid2);
    if (v1 < v2) {
      r = mid2;
    } else
      l = mid1;
  }
  mid1 = solve(v, (mid1 + mid2) / 2);
  printf("%.15lf\n", (double)mid1);
  return 0;
}