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 = 2e5 + 100;
int n, k, x;
int a[maxN];
double s[maxN];
double tim() {
  double res = s[1], sum = 0;
  for (int i = 1; i <= n; i++) {
    sum += s[i];
    res = max(res, sum);
    if (sum < 0) sum = 0;
  }
  return res;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%i", &a[i]);
  double res = 1e18;
  double l = 1e9, r = -1e9;
  for (int i = 1; i <= n; i++)
    l = min(l, double(a[i])), r = max(r, double(a[i]));
  int dem = 100;
  while (dem--) {
    double mid = (l + r) / 2;
    for (int i = 1; i <= n; i++) s[i] = double(a[i]) - mid;
    double s1 = tim();
    for (int i = 1; i <= n; i++) s[i] = -s[i];
    double s2 = tim();
    res = min(max(abs(s1), abs(s2)), res);
    if (abs(s1) > abs(s2))
      l = mid;
    else
      r = mid;
  }
  printf("%.9f", res);
  return 0;
}