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 double eps = 1e-8;
int a[200005], n;
double b[200005];
int read() {
  int x = 0, w = 1;
  char ch = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * w;
}
double solve(double x) {
  double sum = 0, ans = 0;
  for (int i = 1; i <= n; i++) b[i] = a[i] - x;
  for (int i = 1; i <= n; i++) {
    sum += b[i];
    if (sum < 0) sum = 0;
    ans = max(sum, ans);
    b[i] = -b[i];
  }
  sum = 0;
  for (int i = 1; i <= n; i++) {
    sum += b[i];
    if (sum < 0) sum = 0;
    ans = max(sum, ans);
  }
  return ans;
}
int main() {
  int i, j;
  n = read();
  for (i = 1; i <= n; i++) a[i] = read();
  double l = -10000, r = 10000;
  int tt = 100;
  while (tt--) {
    double f1, f2;
    f1 = l + (r - l) / 3, f2 = r - (r - l) / 3;
    double k1 = solve(f1), k2 = solve(f2);
    if (k1 > k2)
      l = f1;
    else
      r = f2;
  }
  printf("%.15f\n", solve(l));
  return 0;
}