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[222222], b[222222];
int n;
double f(double x) {
  for (int i = 0; i < n; i++) b[i] = a[i] - x;
  double MAX = -0x3f3f3f3f, ans = 0;
  for (int i = 0; i < n; i++) {
    ans += b[i];
    MAX = max(ans, MAX);
    if (ans < 0) ans = 0;
  }
  double MIN = 0x3f3f3f3f;
  ans = 0;
  for (int i = 0; i < n; i++) {
    ans += b[i];
    MIN = min(ans, MIN);
    if (ans > 0) ans = 0;
  }
  return max(fabs(MIN), fabs(MAX));
}
int main() {
  scanf("%d", &n);
  double l = 0x3f3f3f3f, r = -0x3f3f3f3f;
  for (int i = 0; i < n; i++) {
    scanf("%lf", &a[i]);
    l = min(a[i], l);
    r = max(a[i], r);
  }
  while (1) {
    if (r - l <= 1e-11) break;
    double mid = (l + r) / 2;
    double mmid = (mid + r) / 2;
    if (f(mid) <= f(mmid))
      r = mmid;
    else
      l = mid;
  }
  double goal = (l + r) / 2;
  printf("%.15f\n", f(goal));
  return 0;
}