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 mcs(int n, double a[200005]) {
  double mx = 0, now = 0;
  for (int i = 0; i < n; i++) {
    now += a[i];
    if (now > mx) mx = now;
    if (now < 0) now = 0;
  }
  return mx;
}
int n, a[200005];
double b[200005];
void calc(double m, double& vp, double& vn) {
  for (int i = 0; i < n; i++) b[i] = a[i] - m;
  vp = mcs(n, b);
  for (int i = 0; i < n; i++) b[i] = -1 * b[i];
  vn = mcs(n, b);
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  double L = *min_element(a, a + n);
  double R = *max_element(a, a + n);
  for (int i = 0; i < 100; i++) {
    double m = (L + R) / 2;
    double vp, vn;
    calc(m, vp, vn);
    if (vp > vn)
      L = m;
    else
      R = m;
  }
  double m = (L + R) / 2;
  double vp, vn;
  calc(m, vp, vn);
  printf("%.7lf\n", max(vp, vn));
  return 0;
}