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 <typename T>
inline void mn(T& x, const T& y) {
  if (y < x) x = y;
}
template <typename T>
inline void mx(T& x, const T& y) {
  if (x < y) x = y;
}
template <typename T>
inline int sz(const T& x) {
  return x.size();
}
const int N = 200000;
int a[N];
double p[N + 1];
double f(const int n, const double x) {
  p[0] = 0;
  for (int i = 0; i < (n); ++i) {
    p[i + 1] = p[i] + (a[i] - x);
  }
  double min_prev = 0;
  double max_prev = 0;
  double min_total = numeric_limits<double>::max();
  double max_total = numeric_limits<double>::min();
  for (int i = 1; i <= n; ++i) {
    mn(min_total, p[i] - max_prev);
    mx(max_total, p[i] - min_prev);
    mn(min_prev, p[i]);
    mx(max_prev, p[i]);
  }
  return max(abs(min_total), abs(max_total));
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < (n); ++i) {
    scanf("%d", &a[i]);
  }
  double l = -1e9;
  double r = +1e9;
  for (int iter = 0; iter < (200); ++iter) {
    double ll = l + (r - l) / 3;
    double rr = r - (r - l) / 3;
    if (f(n, ll) < f(n, rr)) {
      r = rr;
    } else {
      l = ll;
    }
  }
  printf("%.12lf\n", f(n, l));
  return 0;
}