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 &a) {
  a = 0;
  char c;
  while (c = getchar(), !isdigit(c))
    ;
  do a = a * 10 + (c ^ 48);
  while (c = getchar(), isdigit(c));
}
template <class T>
void Min(T &a, T b) {
  if (a > b) a = b;
}
template <class T>
void Max(T &a, T b) {
  if (a < b) a = b;
}
const int M = 2e5 + 5;
int A[M], n;
inline long double Fabs(long double x) {
  if (x < 0) return -x;
  return x;
}
long double eps = 1e-15;
inline long double calc(long double x) {
  long double Mx = -1e30;
  long double sum = 0;
  for (int i = 1; i <= n; ++i) {
    sum += A[i] - x;
    Max(Mx, sum);
    if (sum < 0) sum = 0;
  }
  long double Mn = 1e30;
  sum = 0;
  for (int i = 1; i <= n; ++i) {
    sum += A[i] - x;
    Min(Mn, sum);
    if (sum > 0) sum = 0;
  }
  return max(Fabs(Mx), Fabs(Mn));
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) scanf("%d", &A[i]);
  long double l = -20000, r = 20000;
  for (; Fabs(l - r) > eps;) {
    long double x1 = l + (long double)(r - l) / 3,
                x2 = l + (long double)(r - l) / 3 * 2;
    long double t1 = calc(x1), t2 = calc(x2);
    if (t1 > t2)
      l = x1;
    else
      r = x2;
  }
  printf("%.12lf\n", (double)calc(l));
  return 0;
}