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 N = 200010;
int a[N];
long double d[N];
long double val[N];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  long double l = 1e12, r = 1e12;
  l = -l;
  for (int z = 1; z <= 400; z++) {
    long double mid = (l + r) / 2;
    for (int i = 1; i <= n; i++) d[i] = a[i] - mid;
    val[0] = 0;
    long double res1 = 0;
    for (int i = 1; i <= n; i++) {
      val[i] = max(d[i], d[i] + val[i - 1]);
      res1 = max(res1, val[i]);
    }
    for (int i = 1; i <= n; i++) d[i] = mid - a[i];
    long double res2 = 0;
    for (int i = 1; i <= n; i++) {
      val[i] = max(d[i], d[i] + val[i - 1]);
      res2 = max(res2, val[i]);
    }
    if (res2 > res1)
      r = mid;
    else
      l = mid;
  }
  l = (l + r) / 2;
  for (int i = 1; i <= n; i++) d[i] = a[i] - l;
  long double mini = 0, maxi = 0;
  for (int i = 1; i <= n; i++) {
    d[i] += d[i - 1];
    maxi = max(maxi, d[i]);
    mini = min(mini, d[i]);
  }
  double res = maxi - mini;
  printf("%.13f\n", res);
}