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 n, m, cnt;
double x[200010], inf = 1e9, eps = 1e-12;
double cal(double mi) {
  int i, s;
  double s1, s2, mx, mn;
  s1 = s2 = 0;
  mx = -inf;
  mn = inf;
  for (i = 1; i <= n; i++) {
    if (s1 + x[i] - mi < 0)
      s1 = 0;
    else
      s1 += x[i] - mi;
    if (s2 + x[i] - mi > 0)
      s2 = 0;
    else
      s2 += x[i] - mi;
    mx = max(mx, s1);
    mn = min(mn, s2);
  }
  return max(abs(mx), abs(mn));
}
int main() {
  int i, s, a, b, j, k;
  double l, r, m1, m2;
  while (cin >> n) {
    for (i = 1; i <= n; i++) scanf("%lf", &x[i]);
    l = -inf;
    r = inf;
    for (i = 0; i <= 600; i++) {
      m1 = (l + l + r) / 3.0;
      m2 = (l + r + r) / 3.0;
      if (cal(m1) < cal(m2))
        r = m2;
      else
        l = m1;
    }
    printf("%.10lf\n", cal(l));
  }
  return 0;
}