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;
double a[200010], ans, b[200010], c[200010], sum1[200010], sum2[200010];
double f(double x) {
  double max1 = -100000.0, max2 = -100000.0;
  sum1[0] = sum2[0] = 0.0;
  for (int i = 1; i <= n; i++) {
    b[i] = a[i] - x;
    c[i] = -b[i];
  }
  for (int i = 1; i <= n; i++) {
    sum1[i] = max(sum1[i - 1] + b[i], b[i]);
    max1 = max(max1, sum1[i]);
    sum2[i] = max(sum2[i - 1] + c[i], c[i]);
    max2 = max(max2, sum2[i]);
  }
  return max(max1, max2);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lf", &a[i]);
  if (n == 1)
    printf("%.15lf\n", 0);
  else {
    double L = -20000.0, R = 20000.0;
    for (int i = 1; i <= 200; i++) {
      double mid1 = L + (R - L) / 3.0, mid2 = R - (R - L) / 3.0;
      if (f(mid1) < f(mid2)) {
        ans = mid1;
        R = mid2;
      } else {
        ans = mid2;
        L = mid1;
      }
    }
    printf("%.15lf\n", f(ans));
  }
  return 0;
}