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;
int a[200005];
double temp[200005], dp1[200005], dp2[200005];
double solve(double x) {
  for (int i = 1; i <= n; i++) temp[i] = a[i] - x;
  dp1[0] = 0.0, dp2[0] = 0.0;
  double ans1 = -DBL_MAX, ans2 = -DBL_MAX;
  for (int i = 1; i <= n; i++) {
    dp1[i] = max(dp1[i - 1] + temp[i], temp[i]);
    ans1 = max(ans1, dp1[i]);
  }
  for (int i = 1; i <= n; i++) temp[i] *= -1.0;
  for (int i = 1; i <= n; i++) {
    dp2[i] = max(dp2[i - 1] + temp[i], temp[i]);
    ans2 = max(ans2, dp2[i]);
  }
  return max(ans1, ans2);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  double ans = DBL_MAX;
  double l = -10000, r = 10000;
  int iter = 100;
  while (iter--) {
    double mid1 = (2 * l + r) / 3;
    double mid2 = (l + 2 * r) / 3;
    double val1 = solve(mid1);
    double val2 = solve(mid2);
    if (val1 < val2) {
      r = mid2;
      ans = min(ans, val1);
    } else {
      l = mid1;
      ans = min(ans, val2);
    }
  }
  printf("%.8f\n", ans);
  return 0;
}