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;
template <class T>
inline void Max(T &a, T b) {
  if (a < b) a = b;
}
template <class T>
inline void Min(T &a, T b) {
  if (a > b) a = b;
}
int a[200005];
double cal(double first, int n) {
  double s = 0, mi = 0, mx = 0;
  for (int i = 1; i <= n; i++) {
    s += a[i] - first;
    Min(mi, s);
    Max(mx, s);
  }
  return abs(mx - mi);
}
int main() {
  int T, i, j, k, m, n;
  scanf("%d", &n);
  double l = -1e9, r = 1e9;
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  for (int it = 0; it < 100; it++) {
    double x1 = (11 * l + 10 * r) / 21;
    double x2 = (10 * l + 11 * r) / 21;
    double y1 = cal(x1, n), y2 = cal(x2, n);
    if (y1 >= y2)
      l = x1;
    else
      r = x2;
  }
  printf("%.9lf\n", cal(l, n));
  return 0;
}