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;
int n, a[200001];
const long double esl = 1e-15;
double cal(double x) {
  long double t = 0;
  long double ans = 0;
  for (int i = 1; i <= n; i++) {
    if (t + a[i] - x < 0) {
      ans = max(ans, t);
      t = 0;
    } else {
      t += ((long double)a[i] - x);
      ans = max(ans, t);
    }
  }
  ans = max(ans, t);
  long double t1 = 0;
  for (int i = 1; i <= n; i++) {
    if (t1 + a[i] - x > 0) {
      ans = max(ans, -t1);
      t1 = 0;
    } else {
      t1 += ((long double)a[i] - x);
      ans = max(ans, -t1);
    }
  }
  ans = max(ans, -t1);
  return ans;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  long double l = -(1e9) - 1;
  long double r = 1e9 + 1;
  while (abs(r - l) >= esl) {
    long double d = (2 * l + r) / 3;
    long double c = (2 * r + l) / 3;
    if (cal(d) > cal(c)) {
      l = d;
    } else
      r = c;
  }
  cout.precision(20);
  cout << cal(l);
  return 0;
}