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;
double a[200010];
double b[200010];
double l, r, m, ans1, ans2, res1, res2, ans;
int i, n, x;
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &x), a[i] = x + 10000;
  l = 0;
  r = 20000;
  ans = 4000000000000LL;
  for (int t = 1; t <= 100; t++) {
    m = (l + r) / 2;
    for (i = 1; i <= n; i++) a[i] -= m;
    for (i = 1; i <= n; i++) b[i] = (a[i] * (-1));
    ans1 = ans2 = res1 = res2 = 0;
    for (i = 1; i <= n; i++) {
      ans1 += a[i];
      ans2 += b[i];
      res1 = max(res1, ans1);
      res2 = max(res2, ans2);
      if (ans1 < 0) ans1 = 0;
      if (ans2 < 0) ans2 = 0;
    }
    if (res1 < res2) {
      ans = min(ans, res2);
      r = m;
    } else {
      ans = min(ans, res1);
      l = m;
    }
    for (i = 1; i <= n; i++) a[i] += m;
  }
  printf("%.14f\n", ans);
}