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 inf = (int)1e9 + 7;
const long long INF = (long long)1e18 + 7;
const int N = 1234567;
const int M = 1010;
int n;
double a[N], b[N];
void readInput() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
}
double F(double x) {
  double sum = 0;
  double smax = 0, smin = 0;
  double res = 0;
  for (int i = 1; i <= n; i++) {
    b[i] = a[i] - x;
  }
  for (int i = 1; i <= n; i++) {
    sum += b[i];
    smax = max(smax, sum);
    smin = min(smin, sum);
    res = max(res, max(abs(sum - smax), abs(sum - smin)));
  }
  return res;
}
void Solve() {
  double left = -1e9, right = 1e9;
  for (int it = 0; it < 200; it++) {
    double x1 = left + (right - left) / 3.0;
    double x2 = right - (right - left) / 3.0;
    if (F(x1) > F(x2)) {
      left = x1;
    } else {
      right = x2;
    }
  }
  cout << F(left) << '\n';
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(10);
  cout << fixed;
  readInput();
  Solve();
  return 0;
}