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 = 1e8;
const int maxn = 2e5 + 100;
int n, a[maxn];
double check(double x) {
  double minv = 0.0, maxv = 0.0, ret = 0.0;
  for (int i = 1; i <= n; i++) {
    double tmp = a[i] - x;
    maxv += tmp, minv += tmp;
    ret = max(ret, abs(maxv));
    ret = max(ret, abs(minv));
    if (maxv < 0) maxv = 0;
    if (minv > 0) minv = 0;
  }
  return ret;
}
int main() {
  cin >> n;
  int maxv = -inf, minv = inf;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    maxv = max(maxv, a[i]);
    minv = min(minv, a[i]);
  }
  double l = minv, r = maxv, ans;
  for (int i = 1; i <= 80; i++) {
    double tmp = (r - l) / 3.0;
    double x = l + tmp;
    double y = l + tmp * 2;
    if (check(x) < check(y))
      r = y;
    else
      l = x;
  }
  printf("%.10f", check(l));
  return 0;
}