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 int N = 200002;
const double eps = 1e-7;
int n, i, a[N];
double l, r, lm, rm, u, v, b[N], s[N];
double fabs(double x) { return x > 0 ? x : -x; }
double max(double x, double y) { return x > y ? x : y; }
double f(double x) {
  for (i = 1; i <= n; i++) b[i] = a[i] - x, s[i] = s[i - 1] + b[i];
  double Max = 0, Min = 0, ans = 0;
  for (i = 1; i <= n; i++) {
    ans = max(max(Max - s[i], s[i] - Min), ans);
    if (s[i] > Max) Max = s[i];
    if (s[i] < Min) Min = s[i];
  }
  return ans;
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", a + i);
  l = -10000;
  r = 10000;
  while (1) {
    lm = (l * 2 + r) / 3;
    rm = (l + r * 2) / 3;
    u = f(lm);
    v = f(rm);
    if (fabs(u - v) < eps && r - l < eps) break;
    if (u < v)
      r = rm;
    else
      l = lm;
  }
  printf("%.10lf", u < v ? u : v);
}