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;
int A[200005];
int n, k, x;
struct node {
  double l, r;
};
inline node check(double v) {
  double a = 0, b = 0;
  double mx_a = 0, mx_b = 0;
  for (int i = 1; i <= n; i++) {
    a += 1.0 * (A[i] - v);
    if (a < 0) a = 0;
    mx_a = max(mx_a, a);
  }
  for (int i = 1; i <= n; i++) {
    b += 1.0 * (v - A[i]);
    if (b < 0) b = 0;
    mx_b = max(mx_b, b);
  }
  return (node){mx_a, mx_b};
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &A[i]);
  double l = -10000, r = 10000;
  double ans = 0;
  int num = 100;
  while (num--) {
    double mid = 1.0 * (l + r) / 2;
    node res = check(mid);
    if (res.r >= res.l) {
      ans = res.r;
      r = mid;
    } else
      l = mid;
  }
  printf("%lf\n", ans);
  return 0;
}