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 a[200005], n;
double solve(double x) {
  double sum = 0, maxisum = 0, minisum = 0;
  for (int i = 0; i < n; i++) {
    sum += (a[i] - x);
    if (sum < 0) sum = 0;
    maxisum = max(maxisum, sum);
  }
  sum = 0;
  for (int i = 0; i < n; i++) {
    sum += (a[i] - x);
    if (sum > 0) sum = 0;
    minisum = min(minisum, sum);
  }
  return max(maxisum, -minisum);
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  double esq = -1e18, dir = 1e18;
  int iter = 500;
  while (iter--) {
    double jump = (dir - esq) / 3;
    double t1 = esq + jump;
    double t2 = dir - jump;
    double x1 = solve(t1);
    double x2 = solve(t2);
    if (x1 > x2)
      esq = t1;
    else
      dir = t2;
  }
  printf("%.6f\n", solve(esq));
  return 0;
}