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 cal(int n, int *a, double x) {
  double ret = 0, t1 = 0, t2 = 0;
  for (int i = 0; i < n; i++) {
    t1 += a[i] - x;
    t2 += a[i] - x;
    t1 = max(0., t1);
    t2 = min(0., t2);
    ret = max(ret, t1);
    ret = max(ret, -t2);
  }
  return ret;
}
int main() {
  int n;
  scanf("%d", &n);
  static int a[200000];
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
  }
  double l = -10001, r = 10001, ans;
  for (int t = 0; t < 100; t++) {
    double d1 = (2 * l + r) / 3, d2 = (l + 2 * r) / 3;
    double e1 = cal(n, a, d1), e2 = cal(n, a, d2);
    if (e1 <= e2) {
      r = d2;
      ans = e1;
    } else {
      l = d1;
      ans = e2;
    }
  }
  printf("%.9lf\n", ans);
  return 0;
}