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 MAXN = 1e6 + 10;
const int INF = 2000000010;
const long long INFLL = 999999999999999999ll;
const double eps = 1e-11;
int origin[MAXN];
double cop[MAXN];
int n;
double f(double x) {
  for (int i = 1; i <= n; i++) cop[i] = 1.0 * origin[i] - x;
  double maxx = 0;
  double cur = 0;
  for (int i = 1; i <= n; i++) {
    cur = max(0.0, cur + cop[i]);
    maxx = max(maxx, cur);
  }
  double ret = maxx;
  maxx = 0;
  for (int i = 1; i <= n; i++) cop[i] *= -1;
  cur = 0;
  for (int i = 1; i <= n; i++) {
    cur = max(0.0, cur + cop[i]);
    maxx = max(maxx, cur);
  }
  return max(ret, maxx);
}
double ternary(double left, double right) {
  if (right - left < eps) return (left + right) / 2.0;
  double leftThird = (2.0 * left + right) / 3.0;
  double rightThird = (left + 2.0 * right) / 3.0;
  if (f(leftThird) > f(rightThird))
    return ternary(leftThird, right);
  else
    return ternary(left, rightThird);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &origin[i]);
  printf("%lf\n", f(ternary(-20000.0, 20000.0)));
  return 0;
}