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 a[200005];
int n;
double maxi(double x) {
  double min1 = 0, ans = -1000000007;
  a[1] -= x;
  ans = a[1];
  min1 = min(min1, ans);
  for (int i = 2; i <= n; i++) {
    a[i] -= i * x;
    ans = max(ans, a[i] - min1);
    if (a[i] < min1) min1 = a[i];
  }
  return ans;
}
double mini(double x) {
  double max1 = 0, ans = 1000000007;
  ans = a[1];
  max1 = max(max1, ans);
  a[1] += x;
  for (int i = 2; i <= n; i++) {
    ans = min(ans, a[i] - max1);
    if (a[i] > max1) max1 = a[i];
    a[i] += i * x;
  }
  return ans;
}
int main() {
  int i, j, k, t;
  double l, r, m, tol = 0.0000001, x = 1, y = 100;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%lf", &a[i]);
    a[i] += a[i - 1];
  }
  l = -10000, r = 10000;
  t = 100;
  while (t--) {
    m = (l + r) / 2;
    y = maxi(m);
    x = mini(m);
    if (x > 0)
      l = m + tol;
    else if (y < 0)
      r = m - tol;
    else if (abs(x) == abs(y)) {
      printf("%lf\n", abs(x));
      return 0;
    } else if (abs(x) > abs(y))
      r = m - tol;
    else
      l = m + tol;
  }
  printf("%lf\n", abs(y));
  return 0;
}