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>
void rd(T& t) {
  char c;
  t = 0;
  int k = 1;
  while ((c = getchar()) < '-')
    ;
  if (c == '-') k = -1, (c = getchar());
  do t = t * 10 + (c ^ 48);
  while ((c = getchar()) > '/');
  t *= k;
}
template <class T>
void prinf(T t) {
  if (!t) return;
  prinf(t / 10);
  putchar(t % 10 ^ 48);
}
template <class T>
void ptn(T t) {
  if (!t) putchar('0');
  if (t < 0) putchar('-'), t = -t;
  prinf(t);
}
const int maxn = 2e5 + 5;
int n, a[maxn];
double mx, mi;
void solve(double x) {
  double sum = 0;
  for (int i = (1); i <= (n); i++) {
    sum += a[i] - x;
    mx = max(mx, sum);
    if (sum < 0) sum = 0;
  }
  sum = 0;
  for (int i = (1); i <= (n); i++) {
    sum += a[i] - x;
    mi = min(mi, sum);
    if (sum > 0) sum = 0;
  }
}
int main() {
  rd(n);
  for (int i = (1); i <= (n); i++) rd(a[i]);
  double l = -10001, r = 10001;
  for (int ii = (0); ii <= (100); ii++) {
    double mid = (l + r) / 2;
    mx = 0;
    mi = 0;
    solve(mid);
    if (mx > -mi)
      l = mid;
    else
      r = mid;
  }
  solve(l);
  printf("%lf", max(mx, -mi));
  return 0;
}