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 long long int N = 1e6;
const long long int inf = 1e14;
const long long int mod = 1e9 + 7;
const double PI = acos(-1);
long long int powm(long long int a, long long int b) {
  a = a % mod;
  long long int res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
long long int n, a[N];
long double f(long double xn) {
  set<long double> b = {0};
  long double su = 0, ans = 0;
  long double ma = 0, mi = 0;
  for (int i = 0; i < n; i++) {
    su += (long double)a[i] - xn;
    ans = max({ans, abs(su - mi), abs(su - ma)});
    ma = max(ma, su);
    mi = min(mi, su);
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  cout << fixed << std::setprecision(10);
  while (t--) {
    cin >> n;
    long double B = -100000;
    long double A = 100000;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      B = max(B, (long double)a[i]);
      A = min(A, (long double)a[i]);
    }
    for (int i = 0; i < 100; i++) {
      long double m1 = (A * 2 + B) / 3.0;
      long double m2 = (A + 2 * B) / 3.0;
      if (f(m1) > f(m2))
        A = m1;
      else
        B = m2;
    }
    cout << f(A) << '\n';
  }
}