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;
long long gcd(long long a, long long b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
long double ans = 0;
long double check(vector<long long>& mas, long double& x) {
  long long n = mas.size();
  long double mx1 = mas[0] - x, sum = 0, min_sum = 0;
  for (long long r = 0; r < n; r++) {
    sum += mas[r] - x;
    mx1 = max(mx1, sum - min_sum);
    min_sum = min(min_sum, sum);
  }
  long double mx2 = mas[0] - x;
  sum = 0;
  min_sum = 0;
  for (long long r = 0; r < n; r++) {
    sum += mas[r] - x;
    mx2 = min(mx2, sum - min_sum);
    min_sum = max(min_sum, sum);
  }
  return max(abs(mx1), abs(mx2));
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> mas(n);
  for (long long i = 0; i < n; i++) {
    cin >> mas[i];
  }
  long double l = -10001, r = 10001;
  for (long long i = 0; i < 100; i++) {
    long double lm = (2 * l + r) / 3;
    long double rm = (l + 2 * r) / 3;
    long double a = check(mas, lm), b = check(mas, rm);
    if (a > b) {
      l = lm;
      ans = a;
    } else {
      r = rm;
      ans = b;
    }
  }
  cout << fixed << setprecision(20) << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}