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 read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while ('0' <= c && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
int N;
double a[200010], b[200010], f[200010];
double calc(double x) {
  memset(f, 0, sizeof(f));
  for (int i = 1; i <= N; ++i) b[i] = a[i] + x;
  double ans = 0, res = 0;
  for (int i = 1; i <= N; ++i) {
    if (res + b[i] > 0)
      res += b[i];
    else
      res = 0;
    ans = max(ans, res);
  }
  for (int i = 1; i <= N; ++i) b[i] = -b[i];
  res = 0;
  for (int i = 1; i <= N; ++i) {
    if (res + b[i] > 0)
      res += b[i];
    else
      res = 0;
    ans = max(ans, res);
  }
  return ans;
}
int main() {
  N = read();
  for (int i = 1; i <= N; ++i) a[i] = read();
  double l = -20000, r = 20000;
  for (int cnt = 1; cnt <= 200; ++cnt) {
    double mid = (l + r) / 2;
    double q1 = calc(mid - eps), q2 = calc(mid + eps);
    if (q1 > q2)
      l = mid;
    else
      r = mid;
  }
  printf("%.7lf", calc(l));
  return 0;
}