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 N = 2e5 + 5;
const int mod = 1e9 + 7;
const double inf = 1e18;
int n, a[N];
double l, r, b[N];
double first(double vl) {
  for (int i = 1; i <= n; i++) b[i] = 1.0 * a[i] - vl;
  double sum, minsum, ans = -inf;
  sum = minsum = 0;
  for (int i = 1; i <= n; i++) {
    sum += b[i];
    ans = max(ans, fabs(sum - minsum));
    minsum = min(minsum, sum);
  }
  double maxsum = 0;
  sum = 0;
  for (int i = 1; i <= n; i++) {
    sum += b[i];
    ans = max(ans, fabs(sum - maxsum));
    maxsum = max(maxsum, sum);
  }
  return ans;
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  l = -N, r = N;
  for (int i = 0; i < 100; i++) {
    double m1 = l + (r - l) / 3.0;
    double m2 = r - (r - l) / 3.0;
    if (first(m1) >= first(m2))
      l = m1;
    else
      r = m2;
  }
  printf("%.12f", min(first(l), first(r)));
  return 0;
}