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], n;
double judge(double x) {
  double ret = 0, sum = 0;
  for (int i = 0; i < n; ++i) {
    sum += a[i] - x;
    if (sum < 0)
      sum = 0;
    else {
      ret = max(ret, sum);
    }
  }
  sum = 0;
  for (int i = 0; i < n; ++i) {
    sum += x - a[i];
    if (sum < 0)
      sum = 0;
    else {
      ret = max(ret, sum);
    }
  }
  return ret;
}
int main() {
  double l = 10000, r = -10000, mid, mid1, mid2, t, t1, t2;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
    l = min(l, 1.0 * a[i]);
    r = max(r, 1.0 * a[i]);
  }
  double p, q;
  while (r - l >= 2e-12) {
    p = (2 * l + r) / 3;
    q = (2 * r + l) / 3;
    t1 = judge(p);
    t2 = judge(q);
    if (t1 < t2)
      r = q;
    else
      l = p;
  }
  printf("%.7lf", t1);
  return 0;
}