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 int INF = 0x3f3f3f3f;
const double eps = 1e-8;
const int maxn = 2e5 + 5;
inline long long read() {
  long long ans = 0;
  char ch = getchar(), last = ' ';
  while (!isdigit(ch)) {
    last = ch;
    ch = getchar();
  }
  while (isdigit(ch)) {
    ans = ans * 10 + ch - '0';
    ch = getchar();
  }
  if (last == '-') ans = -ans;
  return ans;
}
inline void write(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
int n, a[maxn];
double calc(double x) {
  double Min = 0, Max = 0, ans = 0;
  for (int i = 1; i <= n; ++i) {
    Min = Min + a[i] - x > 0 ? 0 : Min + a[i] - x;
    Max = Max + a[i] - x < 0 ? 0 : Max + a[i] - x;
    ans = max(ans, max(Max, -Min));
  }
  return ans;
}
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  double L = -10000, R = 10000, x;
  for (int i = 1; i <= 100; ++i) {
    double m1 = L + (R - L) / 3.00;
    double m2 = R - (R - L) / 3.00;
    if (calc(m1) <= calc(m2))
      x = m1, R = m2;
    else
      x = m2, L = m1;
  }
  printf("%.8lf\n", calc(x));
  return 0;
}