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>
const double f1 = (double)1 / 3, f2 = (double)2 / 3;
const double INF = 1e9;
int a[200005], n;
inline double max(double a, double b) { return a > b ? a : b; }
double poor(double x) {
  double m = 0, M = 0, sum = INF, suM = -INF;
  double ret;
  for (int i = 0; i < n; i++) {
    double num = a[i] - x;
    if (sum > 0)
      sum = num;
    else
      sum += num;
    if (suM < 0)
      suM = num;
    else
      suM += num;
    ret = max(ret, max(fabs(sum), fabs(suM)));
  }
  return ret;
}
double solve() {
  double r[2] = {-1e4 - 1, 1e4 + 1};
  for (int t = 0; t < 100 && r[0] < r[1]; t++) {
    double v[2], diff = r[1] - r[0];
    v[0] = poor(r[0] + f1 * diff);
    v[1] = poor(r[0] + f2 * diff);
    if (v[0] < v[1])
      r[1] = r[0] + f2 * diff;
    else
      r[0] = r[0] + f1 * diff;
  }
  return poor(r[0]);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  printf("%f\n", solve());
}