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 long long INF = 3000000000;
int n, m;
double a[2000010], dp[2][2000010];
double judge(double mid) {
  double p1 = 0.0, p2 = 0.0, tmp, sum = 0.0;
  for (int i = 1; i <= n; i++) {
    tmp = a[i] - mid;
    p1 = max(p1 + tmp, tmp);
    sum = max(sum, p1);
    tmp = -tmp;
    p2 = max(p2 + tmp, tmp);
    sum = max(sum, p2);
  }
  return sum;
}
int main() {
  scanf("%d", &n);
  double L = 20000.0, R = -20000.0;
  for (int i = 1; i <= n; i++) {
    scanf("%lf", &a[i]);
    L = min(L, a[i]);
    R = max(R, a[i]);
  }
  while (R - L > 1e-11) {
    double Lmid = (L + L + R) / 3.0;
    double Rmid = (L + R + R) / 3.0;
    if (judge(Lmid) >= judge(Rmid))
      L = Lmid;
    else
      R = Rmid;
  }
  printf("%.12lf\n", judge((L + R) / 2));
  return 0;
}