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 = 200000;
int sum[N + 10];
int main() {
  int n, num;
  double ans;
  while (~scanf("%d", &n)) {
    memset(sum, 0, sizeof(sum));
    scanf("%d", &num);
    sum[0] = num;
    for (int i = 1; i < n; i++) {
      scanf("%d", &num);
      sum[i] = sum[i - 1] + num;
    }
    double left, right, midL, midR;
    double a, b, c, d;
    double flag;
    left = -10000, right = 10000;
    int Z = 110;
    while (Z--) {
      midL = (right - left) / 3 + left;
      midR = (right - left) * 2 / 3 + left;
      a = b = c = d = 0;
      for (int i = 0; i < n; i++) {
        flag = sum[i] + midL * (i + 1);
        a = max(a, flag);
        b = min(b, flag);
        flag = sum[i] + midR * (i + 1);
        c = max(c, flag);
        d = min(d, flag);
      }
      if (a - b > c - d) {
        left = midL;
        ans = c - d;
      } else {
        right = midR;
        ans = a - b;
      }
    }
    printf("%.15lf\n", ans);
  }
  return 0;
}