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 = 2000000;
const double esp = 3e-12;
int n;
double a[N + 5], b[N + 5];
double get_max() {
  double tmp = 0, ans = 0;
  for (int i = 1; i <= n; i++) {
    tmp += b[i];
    if (tmp < 0) {
      tmp = 0;
    }
    ans = max(ans, tmp);
  }
  return ans;
}
double get_ans(double x) {
  double ans1, ans2;
  for (int i = 1; i <= n; i++) b[i] = a[i] - x;
  ans1 = get_max();
  for (int i = 1; i <= n; i++) b[i] = -b[i];
  ans2 = get_max();
  return max(ans1, ans2);
}
double solve() {
  double l = -1e4, r = 1e4;
  while (l + esp < r) {
    double midl = (2 * l + r) / 3.0;
    double midr = (r * 2 + l) / 3.0;
    double w1 = get_ans(midl);
    double w2 = get_ans(midr);
    if (w1 < w2)
      r = midr;
    else
      l = midl;
  }
  return get_ans(l);
}
int main() {
  while (scanf("%d", &n) != EOF) {
    for (int i = 1; i <= n; i++) scanf("%lf", &a[i]);
    printf("%.15f\n", solve());
  }
  return 0;
}