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;
void Rd(int &res) {
  res = 0;
  char p;
  int k = 1;
  while (p = getchar(), !(p >= '0' && p <= '9') && p != '-')
    ;
  if (p == '-') k = -1, p = getchar();
  do {
    res = (res << 1) + (res << 3) + (p ^ 48);
  } while (p = getchar(), p >= '0');
  res *= k;
}
int a[200005], n;
bool gre(double a, double b) { return a - b > 0.0000000001; }
double calc(double pl) {
  double Mx = 0, Mi = 0, Mxsum = 0, Misum = 0, bas = 0;
  for (int i = 1; i <= n; i++) {
    if (gre(bas, Mxsum)) Mxsum = bas;
    if (gre(bas, Misum)) Misum = bas;
    Mxsum += a[i] - pl;
    Misum -= a[i] - pl;
    if (gre(Mxsum, Mx)) Mx = Mxsum;
    if (gre(Misum, Mi)) Mi = Misum;
  }
  if (gre(Mi, Mx))
    return -1;
  else
    return Mx;
}
int main() {
  Rd(n);
  for (int i = 1; i <= n; i++) Rd(a[i]);
  double l = -10000, r = 10000, ans = 0;
  for (int i = 1; i <= 100; i++) {
    double mid = (l + r) / 2;
    double lp = calc(mid);
    if (gre(lp, 0))
      ans = lp, l = mid;
    else
      r = mid;
  }
  printf("%.15lf\n", ans);
  return 0;
}