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 = 2e5 + 10;
int n;
int a[N];
double maxSuff[N];
double minSuff[N];
double mx[N], mn[N];
double f(double x) {
  double tmp;
  for (int i = 1; i <= n; i++) {
    tmp = a[i] - x;
    mx[i] = mn[i] = maxSuff[i] = minSuff[i] = 0;
    maxSuff[i] = max(0.0, maxSuff[i - 1] + tmp);
    minSuff[i] = min(0.0, minSuff[i - 1] + tmp);
    mx[i] = max(mx[i - 1], maxSuff[i - 1] + tmp);
    mn[i] = min(mn[i - 1], minSuff[i - 1] + tmp);
  }
  return max(abs(mx[n]), abs(mn[n]));
}
double tsearch(double l, double r) {
  double a, b;
  for (int i = 0; i < 500; i++) {
    a = l + (r - l) / 3;
    b = r - (r - l) / 3;
    if (f(a) < f(b)) {
      r = b;
    } else {
      l = a;
    }
  }
  return (r + l) / 2;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  printf("%.8lf\n", f(tsearch(-2e9 - 9, 2e9 + 7)));
  return 0;
}