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>
int a[200000];
double diff[200000];
double getAns(int n, double x) {
  for (int i = 0; i < n; i++) {
    diff[i] = a[i] - x;
  }
  double small = 0, big = 0, min = 0, max = 0;
  for (int i = 0; i < n; i++) {
    small += diff[i];
    min = std::min(min, small);
    if (small > 0) small = 0;
    big += diff[i];
    max = std::max(max, big);
    if (big < 0) big = 0;
  }
  return std::max(max, -min);
}
double trinarySearch(int n, double left, double right) {
  int t = 100;
  while (t--) {
    double mid1 = left + (right - left) / 3;
    double mid2 = left + 2 * (right - left) / 3;
    double ans1 = getAns(n, mid1);
    double ans2 = getAns(n, mid2);
    if (ans1 == ans2) {
      left = mid1;
      right = mid2;
    } else if (ans1 < ans2) {
      right = mid2;
    } else {
      left = mid1;
    }
  }
  return (left + right) / 2;
}
int main() {
  int n;
  scanf("%d", &n);
  int max = -10000, min = 10000;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (max < a[i]) max = a[i];
    if (min > a[i]) min = a[i];
  }
  double x = trinarySearch(n, min, max);
  printf("%.10f\n", getAns(n, x));
  return 0;
}