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 int maxn = 2e5 + 10;
int a[maxn];
double b[maxn];
int n;
double maxsum(double a[maxn]) {
  double mx = 0, now = 0;
  for (int i = 1; i <= n; ++i) {
    now += a[i];
    if (now > mx) mx = now;
    if (now < 0) {
      now = 0;
    }
  }
  return mx;
}
double f(double x) {
  for (int i = 1; i <= n; ++i) {
    b[i] = a[i] - x;
  }
  double psum = maxsum(b);
  for (int i = 1; i <= n; ++i) {
    b[i] *= -1;
  }
  double nsum = maxsum(b);
  return max(psum, nsum);
}
double ts(double l, double r) {
  for (int i = 1; i <= 300; ++i) {
    double m1 = l + (r - l) / 3;
    double m2 = r - (r - l) / 3;
    double f1 = f(m1);
    double f2 = f(m2);
    if (f1 > f2) {
      l = m1;
    } else {
      r = m2;
    }
  }
  return f(l);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  double l = *min_element(a + 1, a + n + 1);
  double r = *max_element(a + 1, a + n + 1);
  cout << fixed << setprecision(12) << ts(l, r) << "\n";
  return 0;
}