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>
double a[2000005];
double f(double x, int n) {
  int i;
  double sum[200005] = {0}, maxl = 0, minl = 0;
  for (i = 1; i <= n; i++) {
    sum[i] = sum[i - 1] + a[i - 1] - x;
    if (maxl < sum[i]) maxl = sum[i];
    if (minl > sum[i]) minl = sum[i];
  }
  return maxl - minl;
}
double solve(double l, double r, int n) {
  double mid, mid2;
  int t = 200;
  while (t--) {
    mid = (l + r) / 2.0;
    mid2 = (mid + r) / 2.0;
    if (f(mid, n) < f(mid2, n))
      r = mid2;
    else
      l = mid;
  }
  return l;
}
int main() {
  int n, i;
  double hzb;
  while (~scanf("%d", &n)) {
    for (i = 0; i < n; i++) scanf("%lf", &a[i]);
    hzb = solve(-1e9, 1e9, n);
    printf("%.15lf\n", f(hzb, n));
  }
  return 0;
}