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;
int n, a[233333];
long double g = asin(1) / 5;
long double f(long double x) {
  long double cm = 0, maxm = -2000000001, minm = 2000000001;
  for (int i = 1; i <= n; i++) {
    long double A = a[i] + x;
    if (cm <= 0) cm = 0;
    cm += A;
    maxm = max(maxm, cm);
  }
  cm = 0;
  for (int i = 1; i <= n; i++) {
    long double A = a[i] + x;
    if (cm >= 0) cm = 0;
    cm += A;
    minm = min(minm, cm);
  }
  cm = 0;
  return max(max(maxm, -minm), cm);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  long double l = -20000.0, r = 20000.0;
  for (int i = 1; i <= 400; i++) {
    long double ml = r - l, m1 = l + ml * g, m2 = r - ml * g;
    long double x1 = f(m1), x2 = f(m2);
    if (x1 > x2)
      l = m1;
    else
      r = m2;
  }
  long double g = f(l);
  double ans = g;
  printf("%.6lf\n", ans);
}