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[200010];
double w[200010];
int N;
double check(double x) {
  double ans = 0, tmp = 0;
  for (int i = 0; i < N; i++) {
    if (tmp < 0) tmp = 0;
    tmp += (a[i] - x);
    ans = max(ans, fabs(tmp));
  }
  tmp = 0;
  for (int i = 0; i < N; i++) {
    if (tmp > 0) tmp = 0;
    tmp += (a[i] - x);
    ans = max(ans, fabs(tmp));
  }
  return ans;
}
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) {
    scanf("%lf", &a[i]);
  }
  double l = -1e4, r = 1e4;
  for (int i = 0; i < 100; i++) {
    double d = (r - l) / 3;
    double m1 = l + d, m2 = r - d;
    double c1 = check(m1), c2 = check(m2);
    if (c1 > c2)
      l = m1;
    else
      r = m2;
  }
  printf("%.9f\n", check((l + r) / 2));
  return 0;
}