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 a[200005];
int n;
double calc(double x) {
  double ans1 = 0, sum1 = 0;
  for (int i = 1; i <= n; i++) {
    if (sum1 > 0)
      sum1 += a[i] - x;
    else
      sum1 = a[i] - x;
    ans1 = max(ans1, sum1);
  }
  double ans2 = 0, sum2 = 0;
  for (int i = 1; i <= n; i++) {
    if (sum2 < 0)
      sum2 += a[i] - x;
    else
      sum2 = a[i] - x;
    ans2 = min(ans2, sum2);
  }
  return max(ans1, abs(ans2));
}
int main() {
  int maxval = 0, minval = 20000;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
    maxval = max(maxval, a[i]);
    minval = min(minval, a[i]);
  }
  double l = minval, r = maxval, mid1, mid2;
  for (int times = 0; times < 100; times++) {
    mid1 = (2 * l + r) / 3;
    mid2 = (l + 2 * r) / 3;
    if (calc(mid1) < calc(mid2))
      r = mid2;
    else
      l = mid1;
  }
  printf("%.7lf\n", calc(mid1));
  return 0;
}