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;
const double INF = 1e10;
const int IT = 200;
const int N = 200100;
int n;
double a[N];
double b[N];
double calcMax() {
  double res = 0;
  double bal = 0;
  for (int i = 0; i < n; i++) {
    res = max(res, bal);
    bal += b[i];
    bal = max(bal, 0.);
  }
  return max(res, bal);
}
double calcMin() {
  double res = 0;
  double bal = 0;
  for (int i = 0; i < n; i++) {
    res = min(res, bal);
    bal += b[i];
    bal = min(bal, 0.);
  }
  return min(res, bal);
}
bool check(double x) {
  for (int i = 0; i < n; i++) b[i] = a[i] - x;
  double A = calcMax();
  double B = calcMin();
  return (-B > A);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%lf", &a[i]);
  double L = -INF, R = INF;
  for (int it = 0; it < IT; it++) {
    double x = (L + R) / 2;
    if (check(x))
      R = x;
    else
      L = x;
  }
  for (int i = 0; i < n; i++) b[i] = a[i] - L;
  printf("%.12lf\n", calcMax());
  return 0;
}