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 N = 2e5 + 7;
const int M = 4;
const int inf = 1e9 + 7;
const long long base = 1e18;
const double pi = acos(-1);
const double ep = 1e-9;
int n;
int a[N];
double F[N];
double f(double x) {
  F[0] = 0;
  double mn = 0, mx = 0;
  double ans = 0;
  for (int i = 1; i <= n; i++) {
    F[i] = F[i - 1] + a[i] - x;
    ans = max(ans, max(abs(F[i] - mn), abs(F[i] - mx)));
    mn = min(mn, F[i]);
    mx = max(mx, F[i]);
  }
  return ans;
}
int main() {
  cin >> n;
  for (int i = (int)1; i < (int)n + 1; i++) scanf("%d", a + i);
  double lo = -10000, hi = 10000, g, res;
  for (int t = (int)0; t < (int)100; t++) {
    double g1 = (2 * lo + hi) / 3;
    double g2 = (lo + 2 * hi) / 3;
    if (f(g1) < f(g2))
      hi = g2;
    else
      lo = g1;
  }
  printf("%.9lf", f(lo));
}