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[200011];
int n;
double check(double x) {
  for (int i = 1; i <= n; i++) {
    a[i] -= x;
  }
  double mi = min(0.0, a[1]);
  double ans = 0;
  double ma = max(0.0, a[1]);
  double cur = 0;
  for (int i = 1; i <= n; i++) {
    cur += a[i];
    mi = min(mi, cur);
    ma = max(ma, cur);
    ans = max(ans, cur);
    ans = max(ans, fabs(cur - mi));
    ans = max(ans, fabs(cur - ma));
  }
  for (int i = 1; i <= n; i++) {
    a[i] += x;
  }
  return ans;
}
int main() {
  double lo = 0;
  double hi = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    hi = max(hi, abs(a[i]));
  }
  lo = -hi;
  double mid;
  double m1, m2, v1, v2;
  for (int g = 0; g <= 200; g++) {
    m1 = lo + (hi - lo) / 3;
    m2 = lo + 2 * (hi - lo) / 3;
    v1 = check(m1);
    v2 = check(m2);
    if (v1 > v2) {
      lo = m1;
    } else if (v1 < v2) {
      hi = m2;
    } else {
      lo = m1;
      hi = m2;
    }
  }
  printf("%.10f", check(lo));
}