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;
int a[200000 + 10], ans = -0x7fffffff;
double l[200000 + 10];
void Read(int &x) {
  char c;
  bool f = 0;
  while (c = getchar(), c != EOF) {
    if (c == '-') f = 1;
    if (c >= '0' && c <= '9') {
      x = c - '0';
      while (c = getchar(), c >= '0' && c <= '9') x = x * 10 + c - '0';
      if (f) x = -x;
      ungetc(c, stdin);
      return;
    }
  }
}
double get_poor(double x) {
  int i;
  double sum = 0, ret = 0;
  for (i = 1; i <= n; i++) l[i] = a[i] - x;
  for (i = 1; i <= n; i++) {
    sum += l[i];
    if (sum < 0) sum = 0;
    ret = max(sum, ret);
    l[i] = -l[i];
  }
  sum = 0;
  for (i = 1; i <= n; i++) {
    sum += l[i];
    if (sum < 0) sum = 0;
    ret = max(sum, ret);
  }
  return ret;
}
int main() {
  Read(n);
  int i;
  for (i = 1; i <= n; i++) Read(a[i]);
  double l = -1e4, r = 1e4, m1, m2, v1, v2;
  i = 100;
  while (i--) {
    m1 = l + (r - l) / 3;
    m2 = r - (r - l) / 3;
    v1 = get_poor(m1);
    v2 = get_poor(m2);
    if (v1 > v2)
      l = m1;
    else
      r = m2;
  }
  printf("%.15lf\n", get_poor(l));
}